SOLUTION: Please help me solve this:
A ball is thrown vertically upward from the top of a building 32 feet tall with an initial velocity of 16 feet per second. The distance s (in feet) of
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A ball is thrown vertically upward from the top of a building 32 feet tall with an initial velocity of 16 feet per second. The distance s (in feet) of
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Question 394787: Please help me solve this:
A ball is thrown vertically upward from the top of a building 32 feet tall with an initial velocity of 16 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=32+16t-16t^2.
a. after how many seconds does the ball strike the ground?
b. after how many seconds does the ball pass the top of the building on its way down? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! A ball is thrown vertically upward from the top of a building 32 feet tall with an initial velocity of 16 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=32+16t-16t^2.
a. after how many seconds does the ball strike the ground?
set s to zero and solve for t:
s=32+16t-16t^2
0=32+16t-16t^2
0=2+t-t^2
t^2-t-2=0
(t-2)(+1) = 0
t = {-1, 2}
the -1 is an extraneous solution leaving us with:
t = 2 seconds
.
b. after how many seconds does the ball pass the top of the building on its way down?
set s = 32 and solve for t:
s=32+16t-16t^2
32=32+16t-16t^2
0=16t-16t^2
0=t-t^2
0=t(1-t)
t = {0,1}
the 0 represents when it was first thrown
so, 1 second is your answer
