Find the maximum distance above the ground

Question

An object is projected vertically upward from the top of a building with an initial velocity of 400 feet/sec. Its distance in feet above the ground after t seconds is given by the equation s(t)= -16t^2 +400t +100. Find it's maximum distance above the ground.

Answer 1

s(t) = -16t^2 +400t+100

take the derivative and set =0 to find time at maximum height

velocity =0 at max height

v(t)=s'(t) =-32t+400=0

t = 400/32=25/2= 2.5 seconds to reach max height

max height =s(2.5) =-16(2.5)^2 +400(2.5)+100

= -16(6.25)+1000+100

= -100+1100

= 1000 feet