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.
260 exam-style questions on OCR A Level Maths 3.2 Kinematics, covering 3.2.1 Language of kinematics, 3.2.2 Graphs in kinematics, 3.2.3 Displacement-time and velocity-time graphs, 3.2.4 Constant acceleration formulae, 3.2.5 Constant acceleration in two dimensions (A-level only), 3.2.6 Non-uniform acceleration in one dimension (A-level only), 3.2.7 Non-uniform acceleration in two dimensions (A-level only), 3.2.8 Motion under gravity using vectors (A-level only), and 3.2.9 Projectiles (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.