Question

A body is thrown vertically upwards from the top of a tower from point A. It reaches the ground in time t1. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time t2. If it is allowed to fall freely from A, then the time it takes to reach the ground is given by?

Answer 1

Solution:

Let us assume that, body is thrown with speed x. When it is thrown vertically downwards with speed x, it reaches the ground in time t₂.

We know that:

$\implies$ $\boxed{\sf{s = xt_{2} + (\frac{1}{2})g(t_{2})^{2}}}$

When it is thrown vertically upwards from the top of a tower, it reaches the ground in time $\sf{t_{1}}$.

Given:

$\bullet$ Time taken to come at same point = t₁ - t₂

$\bullet$ Time taken to reach top = (t₁ - t₂)/2

$\bullet$ Velocity at top = 0

Now:

$\implies$ 0 = x - g(t₁ - t₂)/2

$\implies$ x = g(t₁ - t₂)/2

$\implies$ s = t₂g(t₁ - t₂)/2 + (1/2)g(t₂)²

$\implies$ s = gt₁t₂/2

$\implies$ s = (1/2)gt²

Equating both, we get:

$\implies$ (1/2)gt² = gt₁t₂/2

$\implies$ t² = t₁t₂

$\implies$ t = √(t₁t₂)

Therefore:

If it is allowed to fall freely from A, then the time it takes to reach the ground is given by $\sf{\sqrt{(t_{1}t_{2})}}$.

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Answered by: Niki Swar, Goa❤️