A ball S S\,S is thrown from a point O O\,O with initial speed V ms−1V \text{ ms}^{-1}V ms−1 at an angle of elevation θ\thetaθ. After traveling a horizontal distance xxx, its height above the level of O O\,O is yyy.
Show that
y=xtanθ−gx22V2cos2θ y = x \tan \theta - \frac{g x^2}{2 V^2 \cos^2 \theta} y=xtanθ−2V2cos2θgx2Given that θ=45∘\theta = 45^\circθ=45∘ and that when x=10x = 10x=10, y=2y = 2y=2, find the speed of S S\,S at the point where x=10x = 10x=10 and y=2y = 2y=2.