SOLUTION: Please help me with this problem: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

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Question 1034544: Please help me with this problem: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) = -16t2 +64t + 80
a. After how many seconds will the ball reach its maximum height?
b. How long will it take before the ball reaches the ground?
c. What is the maximum height of the ball?

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
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 =
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