Derivative Velocity/MPH

Question

A ball is dropped from the top of the Empire State building to the ground below. The height, y, of the ball above the ground (in feet) is given as a function of time, t (in seconds), by

y = 1250 - 16t²

I have already figured out:

Acceleration: -32

Ball hits the ground at: 8.8388 seconds

The ball is going: -282.84 ft/sec

When does the ball hit the ground, and how fast is it going at that time? Give your answer in feet per second and in miles per hour (1 ft/sec = 15/22 mph).

Please help me figure this question out and tell how you got the answer in mph

Answer 1

Where did the 1ft/sec = 15/22 mi/hr come from?

This is from something called dimensional analysis.

1 ft/set (1mi/5280ft) (60sec/1min)(60 min/1hr)

The above shows the steps for converting 1 ft/sec to mi/hr. Take note of the units and how they cancel out.

ft/sec mi/ft sec/min min/hr (ft cancels out, sec cancels out, min cancels out, leaving mi/hr)

The result is (60)(60)/5280 = 15/22 mi/hr. So the work has already been done for you. Since the velocity is -282.84 ft/sec the speed is 282.84 ft/sec.

In mi/hr:

282.84 ft/sec (15/22)mi/hr / 1 ft/sec. In this case the ft/sec cancel out leaving mi/hr.

Another way to think about it: each ft/sec is 15/22 mi/hr, but you have 282.84 of these ft/sec (so just multiply 282.84 (15/22)).