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vector requires both a magnitude and a specified direction."},{"text":"A vector has magnitude only; a scalar has magnitude and direction.","correct":false,"explanation":"This reverses the definitions. A scalar has magnitude only, while a vector has both magnitude and direction."},{"text":"A scalar has magnitude only; a scalar has magnitude and direction.","correct":false,"explanation":"Calling both quantities scalars overlooks the role of direction. A scalar has magnitude only, but a vector has magnitude and direction."},{"text":"A vector has magnitude only; a vector has magnitude and direction.","correct":false,"explanation":"Calling both quantities vectors incorrectly gives direction to the first quantity. A scalar has magnitude only, whereas a vector also has direction."}]},{"id":"32d08875-9b53-4586-b696-7ac6b2986f26","cardType":"basic","points":1,"front":"Which quantity needs a direction to be complete: distance or displacement?","back":"Displacement.","mcqOptions":[{"text":"Displacement.","correct":true,"explanation":"Displacement is a vector, so its description must include both the change in position and the direction of that change."},{"text":"Distance.","correct":false,"explanation":"Distance is a scalar and therefore needs only a magnitude. Displacement is the quantity that also requires a direction to be complete."},{"text":"Both distance and displacement.","correct":false,"explanation":"This treats distance as a vector, but distance has magnitude only. Of the two quantities, only displacement requires a direction."},{"text":"Neither distance nor displacement.","correct":false,"explanation":"This overlooks that displacement is a vector quantity. Distance needs no direction, but displacement must specify the direction of the change in position."}]},{"id":"13bb5f5e-0a70-4f65-ab98-8f260c1f103f","cardType":"basic","points":1,"front":"If an object returns to its starting point, what is its displacement?","back":"Zero.","mcqOptions":[{"text":"Zero.","correct":true,"explanation":"Displacement depends only on the initial and final positions. When these positions coincide, the change in position is zero regardless of distance travelled."},{"text":"The total distance travelled.","correct":false,"explanation":"This confuses displacement with distance travelled. Returning to the starting point can involve a non-zero journey length, but the displacement is zero."},{"text":"Twice the outward distance.","correct":false,"explanation":"Doubling the outward distance may describe the distance for a return journey, not displacement. Coincident initial and final positions give zero displacement."},{"text":"Equal to the journey length.","correct":false,"explanation":"Journey length is the scalar distance travelled, not the vector change in position. Returning to the starting point makes the displacement zero."}]},{"id":"7d03c15d-07b4-4c1a-b17f-3ab009a918d8","cardType":"cloze","points":null,"front":"In a vector arrow, length shows [ ] and the arrowhead shows [ ].","back":"In a vector arrow, length shows magnitude and the arrowhead shows direction.","mcqOptions":[{"text":"In a vector arrow, length shows magnitude and the arrowhead shows direction.","correct":true,"explanation":"The arrow’s length is drawn to a chosen scale to represent vector magnitude, while its arrowhead indicates the vector’s direction."},{"text":"In a vector arrow, length shows direction and the arrowhead shows magnitude.","correct":false,"explanation":"This swaps the two features. The length of a vector arrow represents magnitude, and the arrowhead indicates direction."},{"text":"In a vector arrow, length shows magnitude and the arrowhead shows magnitude.","correct":false,"explanation":"An arrowhead does not show magnitude; it identifies which way the vector points. Length represents magnitude, while the arrowhead represents direction."},{"text":"In a vector arrow, length shows direction and the arrowhead shows direction.","correct":false,"explanation":"Length cannot identify direction because arrows pointing differently can have equal lengths. Length represents magnitude, and the arrowhead represents direction."}]},{"id":"a94299ed-93ba-44a1-b036-2f4df8006764","cardType":"basic","points":1,"front":"What single vector replaces two or more vectors acting together?","back":"The resultant vector.","mcqOptions":[{"text":"The resultant vector.","correct":true,"explanation":"The resultant vector is the single vector whose magnitude and direction produce the same overall effect as all the original vectors combined."},{"text":"The equilibrant vector.","correct":false,"explanation":"The equilibrant is equal and opposite to the resultant, so it would balance the combined vectors rather than replace them with the same effect."},{"text":"The component vector.","correct":false,"explanation":"A component vector is one resolved part of a vector, usually horizontal or vertical. The single replacement for combined vectors is the resultant vector."},{"text":"The scalar sum.","correct":false,"explanation":"Vector addition must account for direction, so magnitudes cannot simply be combined as a scalar sum. The required single vector is the resultant vector."}]},{"id":"a08abf42-7484-462b-8cd5-0c9cd170d23b","cardType":"basic","points":1,"front":"For vector $A$ at angle $\\theta$ above the horizontal, what is $A_x$?","back":"$A_x=A\\cos\\theta$.","mcqOptions":[{"text":"$A_x=A\\cos\\theta$.","correct":true,"explanation":"The horizontal component is adjacent to the angle measured from the horizontal, so cosine gives $A_x/A$, producing $A_x=A\\cos\\theta$."},{"text":"$A_x=A\\sin\\theta$.","correct":false,"explanation":"Using sine selects the component opposite the angle, which is the vertical component here. The horizontal component is $A_x=A\\cos\\theta$."},{"text":"$A_x=A\\tan\\theta$.","correct":false,"explanation":"Tangent gives the ratio of vertical to horizontal components, not a component as a fraction of $A$. The correct expression is $A_x=A\\cos\\theta$."},{"text":"$A_x=\\frac{A}{\\cos\\theta}$.","correct":false,"explanation":"Dividing by cosine treats $A$ as the horizontal component rather than the hypotenuse. Since $A_x/A=\\cos\\theta$, the correct expression is $A_x=A\\cos\\theta$."}]},{"id":"43a00b4c-66c2-45bf-99f3-7a022a4e563a","cardType":"basic","points":1,"front":"For vector $A$ at angle $\\theta$ above the horizontal, what is $A_y$?","back":"$A_y=A\\sin\\theta$.","mcqOptions":[{"text":"$A_y=A\\sin\\theta$.","correct":true,"explanation":"The vertical component is opposite the angle measured from the horizontal, so sine gives $A_y/A$, producing $A_y=A\\sin\\theta$."},{"text":"$A_y=A\\cos\\theta$.","correct":false,"explanation":"Using cosine selects the component adjacent to the angle, which is the horizontal component here. The vertical component is $A_y=A\\sin\\theta$."},{"text":"$A_y=A\\tan\\theta$.","correct":false,"explanation":"Tangent gives the ratio $A_y/A_x$, not the vertical component as a fraction of the full vector. The correct expression is $A_y=A\\sin\\theta$."},{"text":"$A_y=\\frac{A}{\\sin\\theta}$.","correct":false,"explanation":"Dividing by sine treats $A$ as the opposite side rather than the hypotenuse. Since $A_y/A=\\sin\\theta$, the correct expression is $A_y=A\\sin\\theta$."}]},{"id":"71379fc3-2a03-461d-a24f-d5bad458dede","cardType":"basic","points":1,"front":"If a vector’s angle is measured from the vertical, which component uses cosine?","back":"The vertical component.","mcqOptions":[{"text":"The vertical component.","correct":true,"explanation":"When the angle is measured from the vertical, the vertical component is adjacent to that angle. Cosine relates the adjacent component to the vector’s magnitude."},{"text":"The horizontal component.","correct":false,"explanation":"The horizontal component is opposite an angle measured from the vertical, so it uses sine. Cosine is used for the adjacent vertical component."},{"text":"The resultant component.","correct":false,"explanation":"The resultant is the original vector’s full magnitude, not one of its resolved components. Cosine gives the vertical component when the angle is measured vertically."},{"text":"The perpendicular component.","correct":false,"explanation":"The component perpendicular to the vertical reference direction is horizontal and therefore opposite the angle, so it uses sine rather than cosine."}]},{"id":"e96911a9-f864-4576-93ce-48d946e0ff86","cardType":"basic","points":1,"front":"If upwards is positive, what is the vertical acceleration due to gravity?","back":"$a_y=-g$.","mcqOptions":[{"text":"$a_y=-g$.","correct":true,"explanation":"Taking upwards as positive makes the downward gravitational acceleration negative. Its vertical component is therefore $a_y=-g$ throughout the projectile’s flight."},{"text":"$a_y=+g$.","correct":false,"explanation":"This sign would apply if downwards were chosen as positive. With upwards positive, gravity acts in the negative vertical direction, so $a_y=-g$."},{"text":"$a_y=0$.","correct":false,"explanation":"Zero vertical acceleration would mean that no resultant vertical force acts. Gravity provides a constant downward force, so with upwards positive, $a_y=-g$."},{"text":"$a_y=-\\frac{1}{2}g$.","correct":false,"explanation":"The factor of one-half belongs in the displacement term $rac{1}{2}at^2$, not in the acceleration. The vertical acceleration itself is $a_y=-g$."}]},{"id":"1051713e-1a73-48f5-91c8-749e25ddd025","cardType":"basic","points":1,"front":"Near Earth’s surface, what value of $g$ is used unless told otherwise?","back":"$9.81\\ \\text{m s}^{-2}$.","mcqOptions":[{"text":"$9.81\\ \\text{m s}^{-2}$.","correct":true,"explanation":"Near Earth’s surface, the gravitational field strength gives freely falling objects an acceleration of approximately $9.81\\ \\text{m s}^{-2}$, unless another value is specified."},{"text":"$98.1\\ \\text{m s}^{-2}$.","correct":false,"explanation":"Moving the decimal point one place to the right makes the value ten times too large. Near Earth’s surface, the standard value is $9.81\\ \\text{m s}^{-2}$."},{"text":"$0.981\\ \\text{m s}^{-2}$.","correct":false,"explanation":"Moving the decimal point one place to the left makes the value ten times too small. The standard near-surface value is $9.81\\ \\text{m s}^{-2}$."},{"text":"$4.91\\ \\text{m s}^{-2}$.","correct":false,"explanation":"This incorrectly halves gravitational acceleration, often by confusing $g$ with the coefficient in $rac{1}{2}gt^2$. The value of $g$ is $9.81\\ \\text{m s}^{-2}$."}]},{"id":"e9dadf2e-7d08-4d4b-854f-9707d4b4aef3","cardType":"cloze","points":null,"front":"In ideal projectile motion after launch, the only force is [ ] acting [ ].","back":"In ideal projectile motion after launch, the only force is weight acting downwards.","mcqOptions":[{"text":"In ideal projectile motion after launch, the only force is weight acting downwards.","correct":true,"explanation":"Ideal projectile motion neglects air resistance and assumes no continuing propulsion. The projectile is therefore acted on only by its weight, which acts vertically downwards."},{"text":"In ideal projectile motion after launch, the only force is thrust acting forwards.","correct":false,"explanation":"A launch force acts only while the projectile is being accelerated by the launcher, not throughout its flight. After launch, only weight acts, directed downwards."},{"text":"In ideal projectile motion after launch, the only force is weight acting upwards.","correct":false,"explanation":"Weight is the correct force, but its direction is reversed here. Gravity attracts the projectile towards Earth, so weight acts downwards rather than upwards."},{"text":"In ideal projectile motion after launch, the only force is air resistance acting backwards.","correct":false,"explanation":"Ideal projectile motion explicitly neglects air resistance, so drag is absent. The remaining force is the projectile’s weight, acting vertically downwards."}]},{"id":"eee62c8e-39e3-426d-a061-d1d21954826e","cardType":"cloze","points":null,"front":"A projectile’s [ ] is the horizontal distance travelled before [ ].","back":"A projectile’s range is the horizontal distance travelled before landing.","mcqOptions":[{"text":"A projectile’s range is the horizontal distance travelled before landing.","correct":true,"explanation":"Range measures the projectile’s total horizontal distance from its launch position to the point where it lands, rather than its path length or maximum height."},{"text":"A projectile’s trajectory is the horizontal distance travelled before landing.","correct":false,"explanation":"A trajectory is the curved path followed, not a measure of horizontal distance. The horizontal distance travelled before landing is called the range."},{"text":"A projectile’s range is the horizontal distance travelled before reaching its highest point.","correct":false,"explanation":"Range is measured over the complete flight until landing, not merely until the highest point. The apex usually occurs partway through the horizontal journey."},{"text":"A projectile’s displacement is the horizontal distance travelled before launching.","correct":false,"explanation":"Horizontal displacement is a general vector quantity and is not defined as distance before launching. A projectile’s range is its horizontal distance travelled before landing."}]},{"id":"e1b632fd-6462-4629-af70-440de6699af1","cardType":"basic","points":1,"front":"What quantity is shared by the horizontal and vertical motions of a projectile?","back":"Time.","mcqOptions":[{"text":"Time.","correct":true,"explanation":"Horizontal and vertical motion occur simultaneously, so both components have the same elapsed time $t$. Their velocities, accelerations and displacements are treated separately."},{"text":"Displacement.","correct":false,"explanation":"Horizontal and vertical displacements generally differ and are calculated with different equations. The shared quantity linking both component motions is the elapsed time."},{"text":"Velocity.","correct":false,"explanation":"The horizontal velocity remains constant while the vertical velocity changes under gravity, so velocity is not shared. Both motions instead have the same elapsed time."},{"text":"Acceleration.","correct":false,"explanation":"The components do not share acceleration: horizontal acceleration is zero while vertical acceleration is $-g$ when upwards is positive. They share the same time."}]},{"id":"e1af2d79-27d6-4b12-8718-a80c0776e9d0","cardType":"basic","points":1,"front":"With air resistance ignored, what is a projectile’s horizontal acceleration?","back":"$a_x=0$.","mcqOptions":[{"text":"$a_x=0$.","correct":true,"explanation":"With air resistance neglected, no horizontal force acts on the projectile. Newton’s second law therefore gives zero horizontal acceleration, so horizontal velocity remains constant."},{"text":"$a_x=g$.","correct":false,"explanation":"Gravity acts vertically rather than horizontally, so it does not produce a horizontal acceleration. With no air resistance, the correct horizontal acceleration is $a_x=0$."},{"text":"$a_x=-g$.","correct":false,"explanation":"The negative gravitational acceleration belongs to the vertical component when upwards is positive. There is no horizontal resultant force, so $a_x=0$."},{"text":"$a_x=\\frac{1}{2}g$.","correct":false,"explanation":"The factor of one-half appears in constant-acceleration displacement equations, not as a horizontal acceleration. With no horizontal force, $a_x=0$."}]},{"id":"873cc889-0a20-4fc7-87d9-4520445ed6be","cardType":"basic","points":1,"front":"For launch speed $u$ at angle $\\theta$, what is the horizontal displacement after time $t$?","back":"$x=(u\\cos\\theta)t$.","mcqOptions":[{"text":"$x=(u\\cos\\theta)t$.","correct":true,"explanation":"The initial horizontal velocity is $u\\cos\\theta$ and remains constant because horizontal acceleration is zero. Multiplying this velocity by time gives $x=(u\\cos\\theta)t$."},{"text":"$x=(u\\sin\\theta)t$.","correct":false,"explanation":"Using sine selects the vertical component of the launch velocity rather than the horizontal component. Horizontal displacement is found from $x=(u\\cos\\theta)t$."},{"text":"$x=u\\cos\\theta$.","correct":false,"explanation":"The factor of $t$ is required because displacement equals constant horizontal velocity multiplied by elapsed time. The correct expression is $x=(u\\cos\\theta)t$."},{"text":"$x=(u\\cos\\theta)t-\\frac{1}{2}gt^2$.","correct":false,"explanation":"The gravitational term belongs to vertical displacement because gravity has no horizontal component. Horizontal motion is uniform, giving $x=(u\\cos\\theta)t$."}]},{"id":"49b13b1c-f17c-44b9-8034-bc83c8a38fe1","cardType":"basic","points":1,"front":"For launch speed $u$ at angle $\\theta$, what is the vertical displacement after time $t$?","back":"$y=(u\\sin\\theta)t-\\frac{1}{2}gt^2$.","mcqOptions":[{"text":"$y=(u\\sin\\theta)t-\\frac{1}{2}gt^2$.","correct":true,"explanation":"The initial vertical velocity is $u\\sin\\theta$, while the vertical acceleration is $-g$ when upwards is positive. Substitution into $s=ut+\\frac{1}{2}at^2$ gives this expression."},{"text":"$y=(u\\cos\\theta)t-\\frac{1}{2}gt^2$.","correct":false,"explanation":"Cosine gives the horizontal component of launch velocity, not the vertical component. The vertical displacement is $y=(u\\sin\\theta)t-\\frac{1}{2}gt^2$."},{"text":"$y=(u\\sin\\theta)t+\\frac{1}{2}gt^2$.","correct":false,"explanation":"This sign treats gravity as an upward acceleration. Taking upwards as positive makes gravitational acceleration negative, so the correct term is $-\\frac{1}{2}gt^2$."},{"text":"$y=(u\\sin\\theta)t-gt^2$.","correct":false,"explanation":"Omitting the factor of one-half misapplies the constant-acceleration displacement equation. Since $s=ut+\\frac{1}{2}at^2$, the gravitational term is $-\\frac{1}{2}gt^2$."}]},{"id":"ccaa8dd7-d702-4965-bccf-9a545b81669a","cardType":"cloze","points":null,"front":"At maximum height, the [ ] velocity is zero, but the [ ] velocity remains.","back":"At maximum height, the vertical velocity is zero, but the horizontal velocity remains.","mcqOptions":[{"text":"At maximum height, the vertical velocity is zero, but the horizontal velocity remains.","correct":true,"explanation":"Gravity acts vertically, so the upward vertical component reduces to zero at the top. With air resistance ignored, no horizontal acceleration acts, so the horizontal component continues."},{"text":"At maximum height, the horizontal velocity is zero, but the vertical velocity remains.","correct":false,"explanation":"The components have been swapped. At the highest point, gravity makes the vertical component zero, while the horizontal component remains unchanged because there is no horizontal acceleration."},{"text":"At maximum height, the vertical velocity is zero, but the vertical velocity remains.","correct":false,"explanation":"This incorrectly treats both components as vertical. Only the vertical component reaches zero at maximum height; the horizontal component remains because horizontal acceleration is zero."},{"text":"At maximum height, the horizontal velocity is zero, but the horizontal velocity remains.","correct":false,"explanation":"Horizontal velocity does not become zero at the top, and the same component cannot both be zero and remain. The vertical component is zero there, while horizontal motion continues."}]},{"id":"a956e0e6-a5f3-499d-baf8-d86795234d31","cardType":"basic","points":1,"front":"What condition must hold for SUVAT equations to be valid?","back":"Acceleration must be constant.","mcqOptions":[{"text":"Acceleration must be constant.","correct":true,"explanation":"SUVAT equations are derived for motion with uniform acceleration. If acceleration changes during the motion, a single set of SUVAT equations cannot describe the whole interval."},{"text":"Velocity must be constant.","correct":false,"explanation":"Constant velocity means zero acceleration, which is only a special case. SUVAT equations allow velocity to change, provided the acceleration remains constant throughout the interval."},{"text":"Displacement must be constant.","correct":false,"explanation":"SUVAT equations describe changing displacement during motion, so it need not remain fixed. The required condition is uniform acceleration over the time interval considered."},{"text":"Acceleration must be zero.","correct":false,"explanation":"Zero acceleration is not required; projectiles, for example, accelerate vertically due to gravity. The equations remain valid whenever acceleration is constant, including a non-zero value."}]},{"id":"9ffd69e7-d0ab-42d1-b41e-400ee9659781","cardType":"basic","points":1,"front":"For projectile motion, how should SUVAT equations be applied?","back":"Apply them separately in horizontal and vertical directions.","mcqOptions":[{"text":"Apply them separately in horizontal and vertical directions.","correct":true,"explanation":"Projectile motion is resolved into perpendicular components. SUVAT equations can then be used independently because horizontal and vertical motions have separate accelerations and share the same time."},{"text":"Apply them together using the resultant velocity.","correct":false,"explanation":"Combining the components into a resultant loses the separate acceleration information. The horizontal and vertical equations must be treated independently, while using one common elapsed time."},{"text":"Apply them only in the horizontal direction.","correct":false,"explanation":"The vertical component changes because gravity acts downward, so ignoring it misses the projectile’s fall and flight time. Both components require separate analysis."},{"text":"Apply them only in the vertical direction.","correct":false,"explanation":"Vertical equations determine the fall, but horizontal motion determines quantities such as range. Projectile motion requires separate equations in both directions, not just vertically."}]},{"id":"451cda0d-b078-423b-87cd-95ad082696c9","cardType":"basic","points":2,"front":"When are same-height projectile shortcuts valid?","back":"

Launch and landing heights match.
Air resistance is ignored.

","mcqOptions":[{"text":"

Launch and landing heights match.
Air resistance is ignored.

","correct":true,"explanation":"These shortcuts are derived for a projectile returning to its launch level with no drag. Matching heights gives the symmetric vertical motion required by the formulae."},{"text":"

Launch and landing heights differ.
Air resistance is ignored.

","correct":false,"explanation":"Different launch and landing heights remove the symmetry assumed by the shortcuts. The standard same-height formulae require matching heights as well as negligible air resistance."},{"text":"

Launch and landing heights match.
Air resistance is included.

","correct":false,"explanation":"Air resistance changes the horizontal and vertical motion, so the ideal shortcuts no longer apply. They assume matching heights and that drag is ignored."},{"text":"

Launch and landing speeds match.
Air resistance is ignored.

","correct":false,"explanation":"Equal launch and landing speeds are not the defining condition for these formulae. The projectile must return to its launch height, with air resistance neglected."}]},{"id":"0de4ad41-41f4-4a8d-a130-53239d959c96","cardType":"basic","points":1,"front":"For a same-height projectile, what is the time of flight shortcut?","back":"$T=\\frac{2u\\sin\\theta}{g}$.","mcqOptions":[{"text":"$T=\\frac{2u\\sin\\theta}{g}$.","correct":true,"explanation":"The initial vertical velocity is $u\\sin\\theta$. For a same-height landing, the total time is twice the time needed to reach the top, giving $T=2u\\sin\\theta/g$."},{"text":"$T=\\frac{u\\sin\\theta}{g}$.","correct":false,"explanation":"This is only the time taken to reach maximum height, not the complete flight. Returning to the original level takes an equal additional time, so the result must be doubled."},{"text":"$T=\\frac{2u\\cos\\theta}{g}$.","correct":false,"explanation":"The vertical component controls flight time, so sine is required when the angle is measured from the horizontal. Cosine represents the horizontal component instead."},{"text":"$T=\\frac{2u\\sin\\theta}{g^2}$.","correct":false,"explanation":"Using $g^2$ gives incorrect dimensions and usually comes from mishandling the acceleration term. Dividing the initial vertical velocity by $g$ produces a time."}]},{"id":"78d8aa7a-76b8-467a-ab95-9a03b1a9f297","cardType":"basic","points":1,"front":"For a same-height projectile, what is the maximum height shortcut?","back":"$H=\\frac{(u\\sin\\theta)^2}{2g}$.","mcqOptions":[{"text":"$H=\\frac{(u\\sin\\theta)^2}{2g}$.","correct":true,"explanation":"Maximum height depends on the initial vertical component. Applying the vertical SUVAT relation with final vertical velocity zero gives the squared component divided by $2g$."},{"text":"$H=\\frac{(u\\sin\\theta)^2}{g}$.","correct":false,"explanation":"The factor of $2$ is missing. From $v^2=u^2+2as$, setting the final vertical velocity to zero gives a denominator of $2g$."},{"text":"$H=\\frac{u^2\\sin\\theta}{2g}$.","correct":false,"explanation":"The entire vertical velocity component must be squared, so the sine factor is squared too. Leaving it as $sin\\theta$ underestimates or miscalculates the height."},{"text":"$H=\\frac{(u\\cos\\theta)^2}{2g}$.","correct":false,"explanation":"Maximum height is determined by vertical speed, which is $u\\sin\\theta$ for an angle measured above the horizontal. The cosine component controls horizontal motion."}]},{"id":"e3a142b2-254a-460a-b2f1-cab738fea50f","cardType":"basic","points":1,"front":"For a same-height projectile, what is the range shortcut?","back":"$R=\\frac{u^2\\sin(2\\theta)}{g}$.","mcqOptions":[{"text":"$R=\\frac{u^2\\sin(2\\theta)}{g}$.","correct":true,"explanation":"Range equals horizontal speed multiplied by flight time. Combining the horizontal component with the same-height flight-time expression simplifies to $u^2\\sin(2\\theta)/g$."},{"text":"$R=\\frac{u^2\\sin\\theta}{g}$.","correct":false,"explanation":"This omits the combined angle dependence from multiplying horizontal speed by flight time. The correct simplification contains the double-angle term $\\sin(2\\theta)$."},{"text":"$R=\\frac{u^2\\sin^2\\theta}{g}$.","correct":false,"explanation":"Squaring the sine term confuses the range expression with a squared component. Combining $\\sin\\theta$ and $\\cos\\theta$ produces $\\sin(2\\theta)$ instead."},{"text":"$R=\\frac{2u^2\\sin\\theta}{g}$.","correct":false,"explanation":"This incorrectly doubles the single-angle term rather than using the identity $2\\sin\\theta\\cos\\theta=\\sin(2\\theta)$. The horizontal component also cannot be omitted."}]},{"id":"d48e0eb3-a071-46df-bad0-b34a14241b5b","cardType":"cloze","points":null,"front":"For a horizontal launch, [ ], so vertical motion starts from [ ].","back":"For a horizontal launch, $u_y=0$, so vertical motion starts from rest.","mcqOptions":[{"text":"For a horizontal launch, $u_y=0$, so vertical motion starts from rest.","correct":true,"explanation":"A horizontal launch has no initial component of velocity in the vertical direction. Gravity then accelerates the object downward from zero initial vertical velocity."},{"text":"For a horizontal launch, $u_x=0$, so vertical motion starts from rest.","correct":false,"explanation":"The horizontal component is not zero in a horizontal launch; it is the launch speed. The vertical component is zero, which is why the vertical motion starts from rest."},{"text":"For a horizontal launch, $u_y=0$, so vertical motion starts from uniform motion.","correct":false,"explanation":"Zero initial vertical velocity means the vertical motion starts from rest, not uniform motion. Gravity immediately produces a downward acceleration, so vertical velocity changes."},{"text":"For a horizontal launch, $u_y=g$, so vertical motion starts from rest.","correct":false,"explanation":"$$g$ is the vertical acceleration, not the initial vertical velocity. For a horizontal launch the initial vertical component is zero, so the motion begins from rest."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Scalars, vectors and projectiles","title":"Scalars, vectors and projectiles Flashcards","topbar":{"backHref":"/a-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"2bbb4894-da54-5b6a-92b5-a7fb537536f0","walled":"$L39"}],"$L3a","$L3b"]