A ball is launched straight up from the ground with an initial velocity of 128 feet per second. 19a. At what time does the ball reach its maximum height? and 19. b Find the maximum height of the ball.

Use the equation of motion under gravity

v = u + at where v = final velocity, u = initial velocity, t = time and a = acceleration due to gravity. At maximum height v = 0. Also u = 128 ft/s, a = - 32 f/s/s and t is to be found. So

0 = 128 - 32t

32t = 128

t = 4 secs < - - - answer to 19a.

19b

v^2 = u^2 + 2as where s = distance:-

0 = 128^2 - 2*32*s

s = 128^2 / 64 = 256 feet