Question

When a body is thrown vertically upwards, from the ground, the time of ascent is \( t_1 \) and the time of descent is \( t_2 \) in the absence of air resistance. Then \( t_1 \) is equal to

• A. \( 2t_2 \)
• B. \( 0.5t_2 \)
• C. \( 0.25t_2 \)
• D. \( t_2 \)
• E. \( 4t_2 \)

Hint:

In vertical motion without air resistance, ascent time = descent time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question deals with projectile motion under gravity in one dimension. The key principle is that in the absence of air resistance, the motion is symmetric. The only force acting on the body throughout its flight (both ascent and descent) is gravity, which causes a constant downward acceleration `g`.
Step 2: Key Formula or Approach:
We can use the equations of motion. Let the initial upward velocity be `u`. 1. Time of ascent (\(t_1\)): At the highest point, the final velocity \(v\) is 0. Using \(v = u + at\), we have \(0 = u - gt_1\), so \(t_1 = u/g\). 2. Time of descent (\(t_2\)): The body falls from the maximum height `h` with an initial velocity of 0. The distance it travels is the same as the height it reached. The final velocity just before hitting the ground will be `u` downwards due to symmetry. Using \(v = u + at\), we have \(-u = 0 - gt_2\), so \(t_2 = u/g\). Alternatively, the displacement is the same magnitude for ascent and descent. Let `h` be the max height. Ascent: \(h = ut_1 - \frac{1}{2}gt_1^2\). Also \(u=gt_1\), so \(h = (gt_1)t_1 - \frac{1}{2}gt_1^2 = \frac{1}{2}gt_1^2\). Descent: \(h = \frac{1}{2}gt_2^2\). By equating the expressions for `h`, we can compare \(t_1\) and \(t_2\).
Step 3: Detailed Explanation:
In the absence of air resistance, the acceleration of the body is always the acceleration due to gravity, `g`, directed downwards.
During Ascent: The body moves upwards against gravity. Its speed decreases, and the time taken to reach the maximum height (where velocity becomes zero) is the time of ascent, \(t_1\).
During Descent: The body falls from the maximum height. Its speed increases, and the time taken to return to the ground is the time of descent, \(t_2\).
Because the acceleration is constant and the displacement is of the same magnitude for both parts of the journey, the motion is perfectly symmetric. The speed at any given height is the same on the way up as it is on the way down. Due to this symmetry, the time taken to go up must be equal to the time taken to come down. Therefore, \(t_1 = t_2\).
Step 4: Final Answer:
The time of ascent \(t_1\) is equal to the time of descent \(t_2\).