max height $v=0$. Use $v^2 = u^2 + 2as$: $0 = 10^2 + 2(-9.8)s$ M1\n- $s = 5.1 \\text{ m}$ (2sf) M1\n- Greatest height = $5.1 + 1.5 = 6.6 \\text{ m}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Uses $v=0$ to find displacement to max height.","code":"M"},{"id":"p2","marks":1,"description":"Calculates displacement $s$.","code":"M"},{"id":"p3","marks":1,"description":"Obtains total height $6.6$.","code":"A"}],"finalAnswers":["6.6"],"generatedAt":"2026-05-12T02:14:06.558Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"suvat_maximum_height","label":"Use a SUVAT equation","steps":[{"type":"text","order":0,"title":"Identify the top point","content":"At the greatest height, the particle has stopped moving upwards for an instant, so its velocity is $v=0$. Take upwards as positive, giving $u=10\\text{ m s}^{-1}$ and $a=-9.8\\text{ m s}^{-2}$.\n\nThe launch point is already $1.5\\text{ m}$ above the ground, so first find the **extra vertical displacement** $s$ from that point.","highlights":[{"text":"projected vertically upwards from a point 1.5 m above the ground with a speed of $10\\text{ms}^{-1}$.","location":"specification"},{"text":"Find the greatest height reached by the particle.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=start}{v^2=u^2+2as}"},{"latex":"\\htmlData{anchor=substitute}{0=10^2+2(-9.8)s}"},{"latex":"\\htmlData{anchor=solve}{s=\\frac{100}{19.6}}"}],"highlights":[{"color":"sky","style":"text-color-with-background","anchor":"substitute"}],"connections":[{"to":"substitute","from":"start","color":"sky","label":"substitute values","style":"solid"},{"to":"solve","from":"substitute","color":"orange","label":"rearrange","style":"solid"}]},"order":1,"title":"Apply the SUVAT equation","description":"Use the displayed SUVAT equation because it links velocity, acceleration and displacement without needing time. Setting the velocity at the top to zero earns the first method mark."},{"type":"text","order":2,"title":"Find the rise","content":"So the particle rises by\n\n$$s=5.102\\ldots\\text{ m}\\approx5.1\\text{ m}$$\n\nto $2$ significant figures. This earns the second method mark. This is the height **above the launch point**, not yet above the ground."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{greatest height}=1.5+\\htmlData{anchor=sum}{5.1}"},{"latex":"\\text{greatest height}=\\htmlData{anchor=answer}{6.6\\text{ m}}"}],"connections":[{"to":"answer","from":"sum","color":"green","label":"add"}]},"order":3,"title":"Add the starting height","description":"Add the initial height above the ground to the upward displacement. This earns the final accuracy mark (A1)."}]}],"generatedAt":"2026-07-21T03:57:18.755Z","modelVersion":"gpt-5.6-terra","explanationVersion":"v3"},"options":[]},{"id":"ee67a87a-6f04-4020-938a-21c0c6f04798","label":"b","content":"Find the time for which the particle is more than 3 m above the ground.","markscheme":"- Height above ground = 3m means displacement $s = 1.5$ from start. M1\n- $1.5 = 10t - 4.9t^2 \\implies 4.9t^2 - 10t + 1.5 = 0$ M1\n- $t = 1.88$ and $t = 0.16$. Time difference = $1.88 - 0.16 = 1.7 \\text{ s}$ A1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Identifies $s=1.5$ as the threshold displacement.","code":"M"},{"id":"p2","marks":1,"description":"Solves quadratic for two values of $t$.","code":"M"},{"id":"p3","marks":1,"description":"Calculates the time interval $1.7$.","code":"A"}],"finalAnswers":["1.7"],"generatedAt":"2026-05-12T02:14:06.558Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"quadratic_formula","label":"Form and solve the quadratic","steps":[{"type":"text","order":0,"title":"Measure from the start","content":"The particle starts $1.5\\text{ m}$ above the ground. To be $3\\text{ m}$ above the ground, its **displacement from the starting point** must be $3-1

Constant Acceleration

Question 10

Constant Acceleration Questions

Constant Acceleration

Question 10

Constant Acceleration Questions