In this question, use g=9.8 m s−2g = 9.8 \text{ m s}^{-2}g=9.8 m s−2.
A projectile is launched from a point on horizontal ground with an initial velocity of 14 m s−114 \text{ m s}^{-1}14 m s−1 at an angle θ\thetaθ above the horizontal.
The projectile reaches a maximum vertical height of HHH metres above the ground.
Show that
H=10sin2θ H = 10 \sin^2 \theta H=10sin2θHence, given that 0∘≤θ≤45∘0^\circ \le \theta \le 45^\circ0∘≤θ≤45∘, find the maximum value of HHH.
A student claims that a projectile with a larger mass will always reach a lower maximum vertical height when launched with the same initial velocity and angle. State whether the student is correct, giving a reason for your answer.
163 exam-style questions on WJEC A Level Maths 4.8 Kinematics (A-level only), covering 4.8.1 Kinematics (A-level only), 4.8.2 Kinematics (A-level only), and 4.8.3 Kinematics (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.