SOLUTION: A rock is thrown from the top of a tall building. The distance, d, in feet, between the rock and the ground t seconds after it is thrown is given by the model
d=-16t^2-2t+524
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d=-16t^2-2t+524
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Question 312082: A rock is thrown from the top of a tall building. The distance, d, in feet, between the rock and the ground t seconds after it is thrown is given by the model
d=-16t^2-2t+524
How long after the rock is thrown is it 321 feet from the ground? Answer by solver91311(24713) (Show Source):
So we are at the top of a 524 foot tall building throwing a rock downward with an initial velocity of 2 feet per second toward the ground.
But we want to know the time when
feet.
Set the two expressions equal to each other because they are both equal to .
Which is to say:
Just solve the quadratic for . Hint: The quadratic factors because 2 times -8 is -16, (-8 times -7) plus (2 times -29)\ = 56 minus 58 = minus 2, and -7 times -29 is 203.
One of the roots is negative -- discard it. You don't care what happened before you threw the rock. The positive root is the number of seconds that will have elapsed when the height of the rock is 321 feet.
