TS EAMCET · Physics · Laws of Motion

Two bodies were thrown simultaneously from the origin: one straight up and the other, at an angle, $60 °$ to the vertical. The initial velocity of each body is equal to $10 m s^{- 1}$ Neglecting the air resistance, the distance between the two bodies after $t = 2 s$ is ( $Use g = 10 m s^{- 2}$ )

A $20 m$
B $20 \sqrt{2} m$
C $5 \sqrt{3} m$
D $30 m$

Verified Solution

Answer & Solution

Correct Answer

(A) $20 m$

Step-by-step Solution

Detailed explanation

In the above case first find out the relative speed of the particle, here, v→A=v0j^ and v→B=v0cosθi^+v0sinθj^, ⇒v→AB=v0j^-v0cosθi^-v0sinθj^ and the relative acceleration, a→AB=-gj^--gj^=0 so relative displacement,…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Two bodies were thrown simultaneously from the origin: one straight up and the other, at an angle, 60° to the vertical. The initial velocity of each body is equal to 10 m s-1 Neglecting the air resistance, the distance between the two bodies after t=2 s is ( Use g=10 m s-2)

Step-by-step Solution

Two bodies were thrown simultaneously from the origin: one straight up and the other, at an angle, $60 °$ to the vertical. The initial velocity of each body is equal to $10 m s^{- 1}$ Neglecting the air resistance, the distance between the two bodies after $t = 2 s$ is ( $Use g = 10 m s^{- 2}$ )