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.
Show that
y=xtanθ−gx22u2cos2θ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} y=xtanθ−2u2cos2θgx2Given that θ=45∘\theta = 45^\circθ=45∘ and that when x=6x = 6x=6, y=2y = 2y=2, find the speed of P P\,P at the point where x=6x = 6x=6 and y=2y = 2y=2.