A golf ball is hit from a point O O\,O on level ground with an initial velocity V ms−1V \text{ ms}^{-1}V ms−1 at an angle of α \alpha\,α to the horizontal. The position of the ball at time t t\,t seconds is given by horizontal displacement x x\,x and vertical displacement y y\,y relative to OOO.
Show that the trajectory of the golf ball is given by the equation
y=xtanα−gx22V2cos2α y = x \tan \alpha - \frac{g x^2}{2 V^2 \cos^2 \alpha} y=xtanα−2V2cos2αgx2Given that α=45∘\alpha = 45^\circα=45∘ and that when the horizontal distance x=12x = 12x=12 m, the height y=3y = 3y=3 m, calculate the speed of the golf ball at the instant it is at the point (12,3)(12, 3)(12,3).
265 exam-style questions on AQA A Level Maths 3.2 Q: Kinematics, covering 3.2.1 Language of kinematics, 3.2.2 Graphs in kinematics, 3.2.3 Constant acceleration formulae, 3.2.4 Calculus in kinematics (A-level only), and 3.2.5 Motion under gravity and projectiles (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.