Question

A body is thrown vertically upwards with an initial velocity \( u \). If \( g \) is the acceleration due to gravity, then the total time taken for the body to return to the starting point is:

• A. \( \frac{u}{g} \)
• B. \( \frac{2u}{g} \)
• C. \( \frac{u}{2g} \)
• D. \( \frac{3u}{g} \)

Hint:

For objects thrown vertically upwards: - The time taken to reach maximum height is \( t = \frac{u}{g} \). - The total time for upward and downward motion is \( T_{\text{total}} = \frac{2u}{g} \).

Answer 1

The Correct Option is B

Step 1: Understanding vertical motion
When a body is thrown vertically upwards with initial velocity $u$, it moves against gravity, slows down, reaches a maximum height, and then starts descending due to gravitational pull.
Using the equation of motion for upward motion:
$v = u - gt$
At the highest point, $v = 0$, so:
$0 = u - gt$
Solving for $t$:
$t = \dfrac{u}{g}$
This gives the time taken to reach the maximum height.


Step 2: Finding the total time
Since the motion is symmetrical, the time taken for the body to fall back to the starting point is also $\dfrac{u}{g}$.
Thus, the total time for the upward and downward motion is:
$T_{\text{total}} = \dfrac{u}{g} + \dfrac{u}{g} = \dfrac{2u}{g}$

Hence, the correct answer is option (2) $\dfrac{2u}{g}$.