Question

A freely falling body, falling from a tower of height h covers a distance of h/3 in the last second of its fall. What is the height of the tower?

Answer 1

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Let total time taken to reach the ground be "T" seconds

Use equation of motion

S = ut + 0.5at²

For total fall

h = (0 × T) + 0.5gT²

h = 0.5gT² ......(1)

For the time "T - 1" seconds

h - h/3 = (0 × T) + [0.5g(T - 1)²]

2h/3 = 0.5g(T - 1)² .......(2)

Divide equations (1) and (2)

h / (2h/3) = 0.5gT² / [0.5g(T - 1)²]

3/2 = [T / (T - 1)]²

√3/√2 = T / (T - 1)

T√3 - √3 = T√2

T(√3 - √2) = √3

T = √3 / (√3 - √2) seconds

Substitute value of T in equation (1)

h = 0.5gT²

= 0.5 × 9.8 m/s² × [√3 / (√3 - √2) s]²

= 4.9 × 3/(√3 - √2)² m

= 14.7 / (0.318)² m

≈ 148 m