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A baseball is thrown straight up in the air at with an initial velocity of 60 feet per second (roughly 40 mph). The height of the baseball as a function of time is then given by s(t) = -16 t^2 + 60t. Determine the instantaneous velocity of the ball after 1, 2 and 3 seconds.
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Find the derivative. Then plug in \(1\), \(2\), and \(3\).
Take the derivative of the equation so: -16(2)=-32t+(60)(1) by using the power rule. so s'(t) = -32t+60. Then plug 1, 2, and 3 into the equation to get the instantaneous velocity for the ball after each second. So I got 28 for 1 second, -4 for 2 seconds, and -36 for 3 seconds.
thank you that really helps.
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