Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is:

4.9 m

9.8 m

19.6 m

24.5 m

Step-by-Step Solution

Analyze Time of Motion:

Let the second body be released at $t=0$ relative to its own motion. We are looking for the separation 2 seconds later, so the second body travels for $t_2 = 2 \text{ s}$ .
The first body was released 1 second before the second body. Therefore, it has been falling for $t_1 = t_2 + 1 = 2 + 1 = 3 \text{ s}$ .

Calculate Distance Covered by First Body ( $h_1$ ): Using the equation $s = ut + \frac{1}{2}at^2$ with $u=0$ and $a=g=9.8 \text{ m/s}^2$ : $h_1 = \frac{1}{2} g t_1^2 = \frac{1}{2} (9.8) (3)^2 = 4.9 \times 9 = 44.1 \text{ m}$
Calculate Distance Covered by Second Body ( $h_2$ ): $h_2 = \frac{1}{2} g t_2^2 = \frac{1}{2} (9.8) (2)^2 = 4.9 \times 4 = 19.6 \text{ m}$
Calculate Separation: $\text{Separation} = h_1 - h_2 = 44.1 \text{ m} - 19.6 \text{ m} = 24.5 \text{ m}$

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