SOLUTION: A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet per second. The distance s (in feet) of the ball from the ground af
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Question 459570: A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=128+112t-16t^2
a) After how many seconds does the ball strike the ground?
b) After how many seconds will the ball pass the top of the building on its way down?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet per second. The distance s (in feet) of the ball from the ground after t seconds is
s(t) = 128+112t-16t^2
----------------------------------
a) After how many seconds does the ball strike the ground?
Let s(t) = 0 and solve for "t":
-16t^2+112t+128 = 0
----
Divide thru by -16 to get:
t - 7t+ 8 = 0
-------------------------
(t-8)(t+1) = 0
Positive solution:
time = 8 seconds
================================
b) After how many seconds will the ball pass the top of the building on its way down?
Let s(t) = 128
Solve: -16t^2+112t+128 = 128
-16t^2+112t = 0
t^2 - 7t = 0
t(t-7) = 0
Positive solution:
---
time = 7 seconds
===================
Cheers,
Stan H.
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Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The distance above the ground when the ball hits the ground is zero, so solve:
for the positive root (it certainly didn't hit the ground before you threw it, right?).
The ball is at 128 feet at time 0 and at the time when the ball passes the top of the building on the way down. So solve:
One of the roots is 0, and we know what that means, so you can discard it. The other root is your answer.
John
My calculator said it, I believe it, that settles it
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