Question

A body is projected vertically upwards. The times corresponding to height $ h $ while ascending and while descending are $ t_1 $ and $ t_2 $, respectively. Then, the velocity of projection will be (take $ g $ as acceleration due to gravity)

• A. \( \sqrt{g h} \)
• B. \( \frac{g (t_1 + t_2)}{2} \)
• C. \( g \sqrt{t_1 t_2} \)
• D. \( \frac{g t_1 t_2}{t_1 + t_2} \)

Hint:

The total time for a body projected vertically upwards is the sum of the time taken to reach the maximum height and the time taken to descend back to the ground. Use the formula \( v_0 = \frac{g(t_1 + t_2)}{2} \) to find the velocity of projection.

Answer 1

The Correct Option is B

To determine the velocity of projection when a body is projected vertically upwards, reaching the same height $h$ at times $t_1$ (while ascending) and $t_2$ (while descending), we start by analyzing the motion. Using the kinematic equation for uniformly accelerated motion, we know that for the upward motion at height $h$, the equation is:

$$h = v_0 t_1 - \frac{1}{2}g t_1^2$$

Here, \( v_0 \) is the velocity of projection. Similarly, for the downward motion when the body descends back to the same height:

$$h = v_0 t_2 - \frac{1}{2}g t_2^2$$

By equating these two expressions for $h$, we find:

$$v_0 t_1 - \frac{1}{2}g t_1^2 = v_0 t_2 - \frac{1}{2}g t_2^2$$

Rearranging terms gives:

$$v_0 (t_1 - t_2) = \frac{1}{2}g (t_1^2 - t_2^2)$$

We know that \( t_1^2 - t_2^2 = (t_1 - t_2)(t_1 + t_2) \), therefore:

$$v_0 (t_1 - t_2) = \frac{1}{2}g (t_1 - t_2)(t_1 + t_2)$$

Cancelling \( t_1 - t_2 \) from both sides, assuming \( t_1 \neq t_2 \), we have:

$$v_0 = \frac{g (t_1 + t_2)}{2}$$

Thus, the velocity of projection is:

$$\boxed{\frac{g (t_1 + t_2)}{2}}$$

Answer 2

For a body projected vertically upwards, the time of ascent (\( t_1 \)) and time of descent (\( t_2 \)) are related to the velocity of projection. 
The total time taken to reach the maximum height and return to the ground is \( t_1 + t_2 \). 
The time for ascent \( t_1 \) and descent \( t_2 \) are related by the following equation for a projectile: \[ t_1 = \frac{v_0}{g} \quad \text{and} \quad t_2 = \frac{v_0}{g} \] where \( v_0 \) is the initial velocity (velocity of projection). The total time for the motion is: \[ t_1 + t_2 = \frac{2v_0}{g} \] 
Thus, the velocity of projection \( v_0 \) can be found as: \[ v_0 = \frac{g (t_1 + t_2)}{2} \] Therefore, the velocity of projection is: \[ \text{(B) } \frac{g (t_1 + t_2)}{2} \]