Suppose your friend Max drops a penny from the top floor of the Empire State Building, \(1250\) feet from the ground. After \(t\) seconds the penny is a distance of \(s(t)=-16t^{2}+1250\) from the ground. You are standing about a block away, \(250\) feet from the base of the building.

a. Find a formula for the angle of elevation \(\alpha (t)\) from the ground at your feet to the height of the penny \(t\) seconds after Max drops it. Multiply by an appropriate constant so that \(\alpha (t)\) is measured in degrees.

b. Find a formula for the rate at which the angle of elevation \(\alpha (t)\) is changing at time \(t\), and use the formula to determine the rate of change of the angle of elevation at the time the penny hits the ground.

It is given that, after \(‘t’\) seconds the penny is a distance of \(s(t)=-16t^{2}+1250\) from the ground.

Now, \(tan x=\frac{perpendicular}{base}\)

\(tan (\alpha (t))=\frac{-16t^{2}+1250}{250}\)

\(\alpha (t)= tan^{-1}\left ( \frac{-16t^{2}+1250}{250} \right )\)

It is known that,

\((\pi)radian=(180)degree\)

\(1 degree=\left ( \frac{\pi}{180} \right )radian\)

Therefore,

\(\alpha (t)= tan^{-1}\left ( \frac{-16t^{2}+1250}{250} \right )\cdot \frac{\pi}{180}\)

When penny hits the ground, then \(s(t)=0\)

\(-16t^{2}+1250=0\)

\(t=8.83\) seconds

This implies that after \(8.83\)seconds penny will hit the ground.

From part (a),

\(\alpha (t)= tan^{-1}\left ( \frac{-16t^{2}+1250}{250} \right )\cdot \frac{\pi}{180}\)

Differentiating with respect to \(‘t’\)

\(\frac{d(\alpha(t))}{dt}=\frac{-4\pi t}{5625\left ( \frac{(1250-16t^{2})^{2}}{62500}+1 \right )}\)

Rate at which angle of elevation \(\alpha (t)\) is changing at the time penny hits the ground.

\(\left (\frac{d(\alpha(t))}{dt} \right )_{t-8.83}=\frac{4 \pi (8.83)}{5625(0+1)}\)

\(\left (\frac{d(\alpha(t))}{dt} \right )_{t-8.83}=\frac{-35.32 \pi}{5625}\)