suppose a ball is thrown straight up into the air, and the height of the ball above the ground is given by the function h(t) = 6 + 37t  16t^2, where h is in feet and t is in seconds. What is the velocity of the ball at time t = 3.2
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1 answer

t = 3.2 seconds I presume
h(t) = 6 + 37t  16t^2
velocity = dh / dt
v = 0 + 37  32 t which is initial velocity  g t where g = 32 ft/s^2 (old text:)
at t = 3.2
v = 37  32 (3.2) = 37  102.4 =  70.4 ft/second
by the way at t = 3.2
h = 6 + 37(3.2)  16(3.2)^2 = 6 + 118.4  163.8 = 39.4 feet
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answered by Anonymous
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