Answer:
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
