A ball is launched straight up in the air from a height of 6 feet. Its velocity (feet/second) t seconds after launch is given by
f(t)=-32t+285 Between 2 seconds and 9 seconds the ball's height changed by feet. Round answer to nearest tenth.)
1 answer:
Step-by-step explanation:
A ball is launched straight up in the air from a height of 6 feet. The velocity as a function of time t is given by :
f(t) = -32 t+285
Height of the ball is :
C is constant. Here the ball is launched from a height of 6 feet. So,
At t = 2 s,
At t = 9 s,
Between 2 s and 9 s, the ball's height changed is : 1275 - 512 = 763 m.
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