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).