In this question, use g=9.8 m s−2g = 9.8 \text{ m s}^{-2}g=9.8 m s−2.
A rescue flare is projected from a point at sea level with an initial speed of 28 m s−128 \text{ m s}^{-1}28 m s−1 at an angle of elevation θ\thetaθ. The flare is modelled as a particle moving freely under gravity.
The flare reaches a maximum vertical height of HHH metres.
Show that
H=40sin2θ H = 40 \sin^2 \theta H=40sin2θHence, given that the launch angle is constrained such that 0∘≤θ≤60∘0^\circ \le \theta \le 60^\circ0∘≤θ≤60∘, determine the maximum possible value of HHH.
A technician suggests that a heavier flare will always reach a lower maximum vertical height when launched with the same initial speed and angle. State whether the technician 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.