SOLUTION: A ball is dropped vertically from the top of a building 450 meters above the ground. The height of the ball above the ground after t seconds is given by
h(t)=-4.9t^2 + 0.96t + 450
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h(t)=-4.9t^2 + 0.96t + 450
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Question 1058068: A ball is dropped vertically from the top of a building 450 meters above the ground. The height of the ball above the ground after t seconds is given by
h(t)=-4.9t^2 + 0.96t + 450
You are standing on the ground 10 meters from the base of the building(assume the ground it level). Express the distance between you and the ball as a function of time.
Expressing it as a function of time is giving me trouble. Solving for the hypotenuse of a right triangle with sides 450m 10m I can do. Relating it to time has me stumped. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A ball is dropped vertically from the top of a building 450 meters above the ground. The height of the ball above the ground after t seconds is given by
h(t)=-4.9t^2 + 0.96t + 450
You are standing on the ground 10 meters from the base of the building(assume the ground it level). Express the distance between you and the ball as a function of time.
Expressing it as a function of time is giving me trouble. Solving for the hypotenuse of a right triangle with sides 450m 10m I can do. Relating it to time has me stumped.
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The distance to the ball is the hypotenuse of the right triangle with sides of 10 meters and the ball's height.
h(t)=-4.9t^2 + 0.96t + 450
The +0.96t means the ball was tossed upward at 0.96 m/sec
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