A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as –32 feet per second squared and approximate your answer to 3 decimal places.

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a(t)=-32
v(t)=-32t+C
v(0)=-32(0)+C
40=C
v(t)=-32t+40
What now?

Anti-Differentiate again to get your position function I suppose :)

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