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)=?
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.
To find the height of the object at the time when the speed is zero, you need to determine the time(s) when the derivative of h(t) with respect to t, represented as h'(t), is equal to zero.
First, we need to calculate h'(t) by differentiating the equation h(t) = -16t^2 + 13t + 1 with respect to t. The derivative of -16t^2 is -32t, the derivative of 13t is 13, and the derivative of 1 is 0 since it's a constant term. Thus, h'(t) is equal to -32t + 13.
Next, set h'(t) equal to zero and solve for t:
-32t + 13 = 0
Adding 32t to both sides:
13 = 32t
Then, divide both sides by 32:
t = 13/32
Now that we have the value of t, we can substitute it back into the original equation h(t) = -16t^2 + 13t + 1 to find the height of the object at that time:
h(t) = -16(13/32)^2 + 13(13/32) + 1
Simplifying the equation gives us the final value of h(t).