Question

A body is dropped from the top of a tower. During the last second of its fall it cover 16/25th of the height of the tower.Calculate height of the tower

Answer 1

Answer:

31 m approximately

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 = 16H/25

Distance travelled in (T - 1) = H - 16H/25 = 9H/25

For the time (T - 1) seconds

9H/25 = 0.5g(T - 1)² ......(2)

Divide (1) and (2)

H / (9H/25) = 0.5gT² / [0.5g(T - 1)²]

25/9 = [T / (T - 1)]²

5/3 = T / (T - 1)

5T - 5 = 3T

2T = 5

T = 5/2 seconds

T = 2.5 seconds

From equation (1)

H = 0.5gT²

= 0.5 × 9.8 m/s² × (2.5 s)²

= 30.625 m

Answer 2

Answer:

31 m approximately

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 = 16H/25

Distance travelled in (T - 1) = H - 16H/25 = 9H/25

For the time (T - 1) seconds

9H/25 = 0.5g(T - 1)² ......(2)

Divide (1) and (2)

H / (9H/25) = 0.5gT² / [0.5g(T - 1)²]

25/9 = [T / (T - 1)]²

5/3 = T / (T - 1)

5T - 5 = 3T

2T = 5

T = 5/2 seconds

T = 2.5 seconds

From equation (1)

H = 0.5gT²

= 0.5 × 9.8 m/s² × (2.5 s)²

= 30.625 m

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