A particle P P\,P is projected from a point O O\,O with velocity Ums−1U\text{ms}^{-1}Ums−1 at an angle of θ∘ \theta^\circ\,θ∘ to the horizontal. When P P\,P has moved a horizontal distance xxx, its height above O O\,O is yyy.
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y=xtanθ−gx22u2cos2θ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} y=xtanθ−2u2cos2θgx2Given that θ=30∘\theta = 30^\circθ=30∘ and that when x=532\displaystyle x = \frac{5\sqrt{3}}{2}x=253, y=0y = 0y=0, find the speed of P P\,P at the point where x=532\displaystyle x = \frac{5\sqrt{3}}{2}x=253 and y=0y = 0y=0.
230 exam-style questions on Edexcel A Level Maths Projectiles, covering 6.1 Horizontal Projection, 6.2 Horizontal and Vertical Components, 6.3 Projection at Any Angle, and 6.4 Projectile Motion Formulae. Each one has a worked solution and a mark scheme showing where the marks go.