Question

A body X is allowed to fall freely from a height of 125 m from the ground and another body Y is thrown at the same instant vertically upwards with a velocity of $50 ms^{-1}$ from the ground. The relative velocity of the body Y with respect to the body X when they cross each other is

• A. $125 ms^{-1}$
• B. $100 ms^{-1}$
• C. $50 ms^{-1}$
• D. $25 ms^{-1}$

Hint:

For any two projectiles in air (ignoring air resistance), the relative motion is always {uniform linear motion} (straight line with constant speed), because relative acceleration is zero.

Answer 1

The Correct Option is C

Step 1: Concept of Relative Velocity under Gravity:
When two bodies are moving exclusively under the influence of gravity, they both possess the same acceleration vector, $\vec{a} = \vec{g}$ (directed downwards). Therefore, the relative acceleration between them is: \[ \vec{a}_{rel} = \vec{a}_Y - \vec{a}_X = \vec{g} - \vec{g} = 0 \] Since the relative acceleration is zero, the relative velocity remains constant throughout the motion.
Step 2: Calculating Initial Relative Velocity:
Let's define the upward direction as positive.

• Body X (Dropped): Initial velocity $u_X = 0$.
• Body Y (Thrown Up): Initial velocity $u_Y = +50 \, ms^{-1}$.

Relative velocity of Y with respect to X ($u_{YX}$) at $t=0$: \[ u_{YX} = u_Y - u_X = 50 - 0 = 50 \, ms^{-1} \]
Step 3: Relative Velocity at Crossing:
Since $\vec{a}_{rel} = 0$, the relative velocity at any time $t$ (including when they cross) is equal to the initial relative velocity. \[ v_{YX}(t) = u_{YX} = 50 \, ms^{-1} \]
Step 4: Verification (Optional):
Time to meet: Relative displacement is $125$ m. Relative speed is $50$ m/s. $t = 125/50 = 2.5$ s. At $t=2.5$ s: $v_X = 0 - 10(2.5) = -25$ m/s. $v_Y = 50 - 10(2.5) = 25$ m/s. $v_{YX} = 25 - (-25) = 50$ m/s.
Step 5: Final Answer:
The relative velocity is $50 \, ms^{-1}$.