1. ## projectile

A particle is projected from the top of a cliff which is 425 ft. above sea level and the angle of projection is 450 to the horizontal.
If the greatest height reached above the point of projection is 200 ft, find the speed of the projection and the time taken to reach this greatest height.
Find when and where the particle strikes the sea.

i have one equation using the formula s=ut-.5gt^2
so
200=usin45t-.5gt^2

i also have another equation -425=usin45t-.5gt^2

Im trying to equate these in some way

2. ## Re: projectile

$\Delta y = 200 = v_0\sin(45) \cdot t - \dfrac{1}{2}gt^2$

at the top of its trajectory, $v_y = 0 \implies v_0\sin(45) - gt = 0 \implies t = \dfrac{v_0\sin(45)}{g}$

substitute for $t$ in the $\Delta y$ equation and solve for $v_0$

3. ## Re: projectile

Thanks! I forgot to use that piece of information. The answer i got is u=✓(800g)