Max Height of Ball

Question

A person standing close to the edge on the top of a ​-foot building throws a baseball vertically upward. The quadratic function given below models the​ ball's height above the​ ground, s(t), in​ feet, t seconds after it was thrown. Complete parts​ (a) and (b).

s(t) = -16t^2 = 64t +150

a. After how many seconds does the ball reach its maximum​ height? What is the maximum​ height?

b. How many seconds does it take until the ball finally hits the​ ground?

Answer 1

s(t) = -16t^2 +64t + 150

s'(t) = -32t +64 = 0

t = 64/32 = 2 seconds to reach maximum height

s(2) = -16(2)^2 + 64(2) + 150 = 214 feet = max height

-16t^2 +64t +150 = 0

8t^2 -32t -75 = 0

t = 32/16 + (1/16)sqr(32^2 +32(75))

t = 2 + 3.66

t = about 5.66 seconds to hit the ground