a ball is throw directly upward at a height of 128 ft with an initial velocity of 64 FPS. at how many seconds will the height be 48 ft

Solution:

the acceleration of gravity g = 32.15 ft/s^2

So the height of the ball is h, the time is t

h = v*t - 0.5*g*t^2

Because the initial height = 128 ft and final height = 48 ft

So h= 48-128 = -80 ft

So

64t - 16.08*t^2 = -80

t = 4.98s

So the time is 4.98 seconds.