Asked by Tom
Randy throw a base ball straight upwards, releasing it 6 ft above the round, and giving it an initial speed of v = 128 ft/sec. Use the formula h = -16t^2 + vt + 6 ft( where t = time , h = height ) to find the maximum height of the ball to reach this height.
Answers
Answered by
Steve
h = -16t^2 + 128t + 6
= 262 - 16(t-4)^2
so h is max at t=4. Plug in t=4 to find the max h.
= 262 - 16(t-4)^2
so h is max at t=4. Plug in t=4 to find the max h.
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