To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of the pole after 0.50 s and then reaches the top of the pole again after a total elapsed time of 4.1 s. How high is the pole above the point where the ball was launched

Explanation:

Given that

The initial time to reach pole t=0.5s

Final time to reach pole again t'=4.1s

Final velocity v=0m/s, at maximum height

Then, let find the initial velocity between the 0.5 sec and maximum height.

The time it spent up above the pole is t=4.1-0.5

t=3.6s

So he spent 3.6 s moving up and coming down to meet the pole again.

Therefore, the time to reach maximum height is 3.6/2= 1.8s

Then the velocity at the top of the pole moving up is given as

Using the equation of motion under free fall.

v=u-gt. Against gravity

0=u-9.8×1.8

0=u-17.64

u=17.64m/s

This is the velocity going up at the top of the pole

Now, let know the original initial velocity from the hand of the Abby

Now the total time is now the first 0.5 sec to reach height op pole plus the 1.8sec to reach maximum height, t=0.5s+1.8=2.3s

Then using same equation

v=u'-gt

0=u'-9.8×2.3

0=u'-22.54

Then, u'=22.52m/s

So to calculate height of pole using any of the equation of motion

v²=u²-2gH

In the case the initial velocity is the velocity at Abby hand (u') and the final velocity at the top of the pole(u)

Then

u²=u'²-2gH

17.64²=22.52²-2×9.81×H

311.17=507.15-19.62H

311.17-507.15=-19.62H

-195.98=-19.62H

Then, H=-195.98/-19.62

H=9.99m

Then, the height of the pole is approximately 10m

an energy transformation flow diagram is shown

where is it ?