need help with college algebra

Question

An astronaut on the moon throws a baseball upward. The astronaut is 6​ ft, 6 in.​ tall, and the initial velocity of the ball is

40

ft per sec. The height s of the ball in feet is given by the equation

s equals negative 2.7 t squared plus 40 t plus 6.5

​,

where t is the number of seconds after the ball was thrown.

Answer 1

s(t) = -2.7t^2 +40t +6.5

take the first derivative, set =0

s'(t) = -5.4t + 40 = 0

5.4t = 40

t = 40/5.4 = 20/2.7 = about 7.4 seconds and the ball reaches its maximum height.'

plug that value into the original equation to solve for the maximum height

s(20/2.7) = -2.7(20/2.7)^2 + 40(20/2.7) + 6.5 = -400/2.7 +800/2.7 + 6.5 = 6.5 + 400/2.7 = about 154.5 feet

for maximum height

to find when it hits the ground let s(t) = 0 and solve for t