Question 1034544
A person standing close to the edge on the top of an 80-foot tower throws a ball with an initial speed of 64 feet per second. After t seconds, the height of the ball above the ground is
s(t) = -16t^2 + 64t + 80
a. After how many seconds will the ball reach its maximum height?
Max height will occur on the axis of symmetry, 
find that using t = -b/(2a) where a=-16 and b=64
t = {{{(-64)/(2*-16)}}}
t = 2 seconds
:
b. How long will it take before the ball reaches the ground?
When the ball hits the ground, s(t) = 0, therefore
-16t^2 + 64t + 80 = 0
Simplify, divide by -16, then we can factor
t^2 - 4t - 5 = 0
(t-5)(t+1) = 0
the positive solution is what we want here
t = 5 sec to hit the ground
:
c. What is the maximum height of the ball?
max height occurs after 2 sec, replace t with 2 in the original equation
s(t) = -16(2^2) + 64(2) + 80
s(t) = -64 + 128 + 80
s(t) = 144 ft is the max height