The speed is zero when the derivative of h with respect to t is zero.
dh/dt = -32t + 13 = 0
Solve that for t, and use the t that you get in the equation for h.
t = 13/32 = 0.4063 seconds @ maximum h
h(@ t = .4063) = ___?
The problem corresponds to an object thrown upward at a speed of 13 ft/s from a height of one foot. It doesn't get very high.
if h(t) represents the height of an object above ground level at time t and h(t) is given by h(t)=-16t^2+13t+1 find the height of the object at the time when the speed is Zero. h(t)=?
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