SOLUTION: A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground af

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Question 61621: A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=112+96t-16t^2. After how many seconds will the ball pass the top of the building on its way down?
I BELIEVE I WOULD USE THE Distance=Rate X Time
Is that right? Is this how I would start it?
112+96t-16t^2=r x t.....i got stuck
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The use of the distance formula, d = rt, is not appropriate for this kind of problem. The reason is the distance formula applies in cases of constant motion (speed) whereas this problem involves acceleration (due to gravity).
You have the appropriate function of distance (really height) as a function of time.
The question here is...when will the ball reach a height (distance) of 112 feet after being thrown upward?
To find out, wou want to set the given function of s(t) equal to 112 ft. and solve for the time, t.
112+%2B+96t+-+16t%5E2+=+112 Subtract 112 from both sides.
96t+-+16t%5E2+=+0 Factor out a t.
t%2896+-+16t%29+=+0 Apply the zero product principle.
t+=+0 and 96+-+16t+=+0
At t = 0 seconds, the ball is at 112 feet because that the initial height of the throw.
96+-+16t+=+0 Add 16t to both sides.
96+=+16t Divide both sides by 16.
t+=+6
The ball will be at 112 feet again in 6 seconds.