In this question use g=9.8g = 9.8g=9.8 m s−2^{-2}−2.
A particle P P\,P is projected from a point O O\,O with speed U U\,U m s−1^{-1}−1 at an angle θ \theta\,θ above the horizontal. Air resistance is modelled as negligible. When P P\,P has moved a horizontal distance x x\,x metres, its height above O O\,O is y y\,y metres.
By writing down expressions for x x\,x and for y y\,y in terms of the time ttt, and eliminating ttt, show that y=xtanθ−gx22U2cos2θy = x\tan\theta - \dfrac{gx^2}{2U^2\cos^2\theta}y=xtanθ−2U2cos2θgx2.
It is given that θ=45°\theta = 45°θ=45°, and that y=2y = 2y=2 when x=6x = 6x=6. Find the speed of P P\,P at the point where x=6x = 6x=6 and y=2y = 2y=2.
91 exam-style questions on OCR (MEI) A Level Maths 3.3 Projectiles (A-level only), covering 3.3.1 Model motion under gravity using vectors (A-level only), 3.3.2 Position, velocity, range and maximum height (A-level only), 3.3.3 Find the initial velocity of a projectile (A-level only), 3.3.4 Equation of the trajectory (A-level only), and 3.3.5 Solve simple projectile problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.