TS EAMCET · Physics · Motion In One Dimension

Two balls are released from the same position at a height of $500 m$ above ground, one after the other, with an interval of $1 s .$ What is the distance between the two balls when the first hits ground?
(Acceleration due to gravity $g = 10 m / s^{2}$ )

A $95 m$
B $65 m$
C $130 m$
D $175 m$

Verified Solution

Answer & Solution

Correct Answer

(A) $95 m$

Step-by-step Solution

Detailed explanation

Time of flight of first ball is, t=2hg ⇒t=2×50010=10 s Distance travelled by first ball is, s1=0+12×g×102=500 m Distance travelled by second ball is, s2=0+12×g×10-12 ⇒s2=0+12×10×92=405 m Seperation between the balls…

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 balls are released from the same position at a height of 500 m above ground, one after the other, with an interval of 1s. What is the distance between the two balls when the first hits ground?(Acceleration due to gravity g=10 m/s2)

Step-by-step Solution

Two balls are released from the same position at a height of $500 m$ above ground, one after the other, with an interval of $1 s .$ What is the distance between the two balls when the first hits ground?
(Acceleration due to gravity $g = 10 m / s^{2}$ )