Question 425432: A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per second from an initial height of 112 feet off the ground, then the height of the projectile, h , in feet, t seconds after it's shot is given by the equation: h= -16t^2+128t+112
Find the two points in time when the object is 147 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per second from an initial height of 112 feet off the ground, then the height of the projectile, h , in feet, t seconds after it's shot is given by the equation: h= -16t^2+128t+112. Find the two points in time when the object is 147 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).
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You can see from this equation that the projectile will follow the path of a parabola that opens downwards. The height of the projectile will be at 147 feet twice during its journey, on its way up and on its way down.
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147=-16t^2+128t+112
-16^t^2+128t+112-147=0
16t^2-128t+35=0
solve by using following quadratic formula:
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a=16, b=-128, c=35
t=[-(-128)+-sqrt(128^2-4*16*35)]2*16
=[128+-sqrt(14144)]/32
=(128+-118.93)/32
=7.72 sec and 0.28 sec
ans:
after 0.28 sec the projectile will reach a height of 147 feet on its way up, and after 7.72 sec it will be at a height of 147 feet on its way down.
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