Physics, 18.02.2020 01:16 Destinywall
A bowling ball is launched off the top of a 240-foot tall building. The height of the bowling ball above the ground t seconds after being launched is s(t)=โ16t2+32t+240s(t)=โ16t2+32t+24 0 feet above the ground. What is the velocity of the ball as it hits the ground?
Answers: 3
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A bowling ball is launched off the top of a 240-foot tall building. The height of the bowling ball a...
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