one question help

Use the constant acceleration formulae.

(a)   Use s = ut + (1/2)at²  where s = displacement in time t, with initial velocity u and acceleration a.

For the vertical direction we have:

 a = -32 ft/s² , u = 70*sin(30°) ft/s  and s = -3 ft  (taking the height when hit as the zero height).

So:   -3 = 70*sin(30°)*t - (1/2)*32*t²    or  16t² - 35t - 3 = 0  (noting that sin(30°) = 1/2)

Solve this for t (using the positive solution of course!).

(b)  For the horizontal direction there is zero acceleration, so:   distance = 70*cos(30°)*t  where t is the value found from part (a).

Here is my answer. This is how we do it in physics.