Question

A man releases a stone (at rest, Vo=0) from the top of a tower. During the last second of its travel, the stone falls through a distance of (9/25)H, where H is the tower's height. Find H. Acceleration due to gravity, g=9.8 m/s². Assume no air resistance.

Answer 1

Answer:

Height of tower equals 122.5 meters.

Explanation:

Since the height of the tower is 'H' the total time of fall of stone 't' is calculated using second equation of kinematics as

Since the distance covered in last 1 second is $\frac{9H}{25}$ and the total distance covered in 't' seconds is 'H' thus the distance covered in the first (t-1) seconds of the motion equals

$S_{t-1}=S_{t}-S_{last}\\\\S_{t-1}=H-\frac{9H}{25}=\frac{16H}{25}$

Now by second equation of kinematics we have

$S=ut+\frac{1}{2}gt^{2}\\\\S=\frac{1}{2}gt^{2}(\because u=0)$

Thus we have

$\frac{16H}{25}=\frac{1}{2}g(t-1)^{2}.............(i)\\\\H=\frac{1}{2}gt^{2}..............(ii)$

Dividing i by ii we get

$\frac{16}{25}=\frac{(t-1)^{2}}{t^2}\\\\\therefore \frac{t-1}{t}=\frac{4}{5}\\\\\therefore t=5secs$

Thus from equation ii we obtain 'H' as

$H=\frac{1}{2}\times 9.8\times 5^{2}=122.5meters$