Bri S.

# A stone is thrown straight up from the edge of a roof, 650 feet above the ground, at a speed of 12 feet per second.

A stone is thrown straight up from the edge of a roof, 650 feet above the ground, at a speed of 12 feet per second.

A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later?

B. At what time does the stone hit the ground?

C. What is the velocity of the stone when it hits the ground?

I am not sure where to start...idk if 12t-32t^2 is my height equation and so 12-32t=0 give mes .375 for t and I plug that value into 12t-3t^2???

By:

Bri S.

that makes sense, thank you MIke!
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08/01/20

Mike D.

tutor
Bri. -2 (8t^2 - 6t -325) = 0 is correct, simplifying to 8t^2 - 6t - 325. It's true that 2t (4t-3) = 325. You can't solve this by setting 2t = 325, and 4t-3 = 325 because you wont get a 325 as the product of the 2 brackets then. You can use the quadratic formula so solve 8 t^2 - 6t - 325 =0, setting a=8 b = -6, c = - 325. Mike
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08/01/20

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