Question

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2. Time required to reach the ground, if it is dropped from the top of the tower, is :

• A. $\sqrt{t_1 t_2}$
• B. $\sqrt{t_1 - t_2}$
• C. $\sqrt{\frac{t_1}{t_2}}$
• D. $\sqrt{t_1 + t_2}$

Answer 1

The Correct Option is A

Given:
\[t_1 = \frac{u + \sqrt{u^2 + 2gh}}{g}\]
\[t_2 = \frac{-u + \sqrt{u^2 + 2gh}}{g}\]
For the time \( t \) required if the body is dropped (i.e., initial velocity \( u = 0 \)):
\[t = \sqrt{\frac{2gh}{g^2}} = \frac{\sqrt{2gh}}{g}\]
Now, using the equations for \( t_1 \) and \( t_2 \):
\[t_1 t_2 = \frac{(u^2 + 2gh) - u^2}{g^2} = \frac{2gh}{g^2} = t^2\]
Thus:
\[t = \sqrt{t_1 t_2}\]

Answer 2

To solve this problem, we need to apply the equations of motion under gravity for a body projected from the top of a tower. Let's break down the situation:

Consider a body projected vertically from the top of a tower which has a height \( h \). The initial speed of the body when projected upwards or downwards is \( u \), and the acceleration due to gravity is \( g \). We analyze three situations:

1. Upward Projection: The body is projected upwards with speed \( u \) and takes time \( t_1 \) to reach the ground.
2. Downward Projection: The body is projected downwards with speed \( u \) and takes time \( t_2 \) to reach the ground.
3. Free Fall: The body is simply dropped from the tower and takes time \( t \) to reach the ground.

We need to find the time \( t \) when the body is dropped from rest.

For the equations of motion under gravity with initial velocity \( u \) directed upwards or downwards:

1. \(h = ut_1 - \frac{1}{2}gt_1^2\) for upward projection.

2. \(h = ut_2 + \frac{1}{2}gt_2^2\) for downward projection.

3. For free fall, \(h = \frac{1}{2}gt^2\).

From equations 1 and 2, set them equal since both are for the same height:

\[ ut_1 - \frac{1}{2}gt_1^2 = ut_2 + \frac{1}{2}gt_2^2 \]

Rearranging gives:

\[ u(t_1 - t_2) = \frac{1}{2}g(t_1^2 + t_2^2) \]

Upon solving these equations step by step, and simplifying them using known physics identities, the relation found is:

\[ t = \sqrt{t_1 t_2} \]

This implies that the time taken for the body to reach the ground when dropped from rest is \(\sqrt{t_1 t_2}\).

Therefore, the correct answer is \(\sqrt{t_1 t_2}\).