#1**+2 **

Use the constant acceleration formulae.

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

For the vertical direction we have:

a = -32 ft/s^{2} , 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*t^{2} or 16t^{2} - 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).

Alan Apr 24, 2020