A projectile is launched from the origin O O\,O with an initial velocity W ms−1W \text{ ms}^{-1}W ms−1 at an angle β \beta\,β above the horizontal. Its position (x,y)(x, y)(x,y) is measured from OOO.
Show that
y=xtanβ−gx22W2cos2β y = x \tan \beta - \frac{g x^2}{2 W^2 \cos^2 \beta} y=xtanβ−2W2cos2βgx2Given that β=60∘\beta = 60^\circβ=60∘ and that when x=15x = 15x=15, y=5y = 5y=5, find the speed of the projectile at the point where x=15x = 15x=15 and y=5y = 5y=5.
