A ball is shot from the ground straight up into the air with initial velocity of 40 ft/sec. Assuming that the air resistance can be ignored, how high does it go?

Answer 1

h=40t-16t^2

v=h'=40-32t=0

t=40/32=5/4=

1.25 seconds to reach max height

h(5/4)=40(5/4)-16(5/4)^2= 50-25=

25 feet= max height