Question

13.
A body falls freely from the top of a tower. It
covers 36% of the total height in the last second
before striking the ground level. The height of
the tower is :
(1) 50m (2) 75m (3) 100m (4) 125m​

Answer 1

Answer:

125 m

Explanation:

Let the height of tower be "H"

Time taken to reach the ground = T

Use equation of motion

• S = ut + 0.5at²

For entire time "T"

H = 0.5gT² ...…..(1)

Distance travelled in last second = 36% of H

= (36/100) × H

= 9H/25

Distance travelled in (T - 1) = H - 9H/25

= 25H/25 - 9H/25

= 16H/25

For the time (T - 1) seconds

16H/25 = 0.5g(T - 1)² ......(2)

Divide equations (1) and (2)

H / (16H/25) = (0.5gT²) / [0.5g(T - 1)²]

25/16 = [T / (T - 1)]²

5/4 = T / (T - 1)

5T - 5 = 4T

T = 5 seconds

From equation (1)

H = 0.5gT²

= 0.5 × 10 m/s² × (5 s)²

= 125 m