An object is thrown upward from the ground with an initial velocity of 32ft/s. What is the maximum height the object obtains using the formula s = -16t^2 + 32t, where s = distance above the ground in feet, and t= time in seconds?

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Answer 1 · 2015-04-14T11:20:54

The maximum height with respect to time will occur when the derivative of the distance(time) function equals $0$

$s = -16t^2+32t$

$(ds)/(dt) = -32t+32$

Maximum occurs when
$-32t+32=0$
$rarr t=1$

When $t=1$ the object is at a height of
$-16(1)^2 +32(1)$

$=16$ (feet)