Aball 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  

Idon’t know what the answer is i wish i could