Question 479546: A ball is thrown vertically upward at a velocity of 15 feet per second from a bridge that is 55 feet above the level of the water. The height h, in feet, of the ball above the water at the time t in seconds after it is thrown is
h = 16t^2 + 20t - 48
Find the time when the ball strikes the water
Found 2 solutions by lwsshak3, Alan3354:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! A ball is thrown vertically upward at a velocity of 15 feet per second from a bridge that is 55 feet above the level of the water. The height h, in feet, of the ball above the water at the time t in seconds after it is thrown is
h = 16t^2 + 20t - 48
Find the time when the ball strikes the water.
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When the ball strikes the water, its height above the water=0
h = 16t^2 + 20t - 48=0
solve by quadratic formula as follows:
..
..
a=16, b=20, c=-48
t=[-20ħsqrt(400-4*16*-48)]2*16
t=[-20ħ√3472]/32
t=(-20ħ58.92)/32
t=1.22 sec
or
t=-2.47 sec (reject,t>0)
ans:
The ball strikes the water 1.22 sec after it is thrown
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! h = 16t^2 + 20t - 48
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If the initial upward speed is 15 ft/sec and the bridge is 55 ft above the water, the eqn would be
h(t) = -16t^2 + 15t + 55
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The equation you gave shows increasing altitude with time. It starts at -48 feet and never comes down.
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