From the question
We are given the equation
[tex]y=-\frac{1}{80}x^2+90[/tex]
The horizontal component of the velocity is a constant of 5ft/s
We are to find the rate of change of its elevation at x = 32
First, we will differentiate the equation with respect to t
This gives
[tex]\frac{dy}{dt}=-\frac{2}{80}x\text{.}\frac{dx}{dt}[/tex]
Since the horizontal component of the velocity is a constant of 5ft/s
Then
[tex]\frac{dx}{dt}=5ft\/s[/tex]
Since x = 32
Hence we have
[tex]\frac{dy}{dt}=-\frac{2}{80}(32)(5)[/tex]
Simplifying this we get
[tex]\frac{dy}{dt}=4[/tex]
Therefore, the rate of change of its elevation is 4ft/s
