TS EAMCET · Physics · Motion In One Dimension
Ball-1 is dropped from the top of a building from rest. At the same moment, ball- 2 is thrown upward towards ball-l with a speed \(14 \mathrm{~m} / \mathrm{s}\) from a point \(21 \mathrm{~m}\) below the top of building. How far will the ball-1 have dropped when it passes ball-2. (Assume, acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\).)
- A \(\frac{45}{4} \mathrm{~m}\)
- B \(\frac{52}{6} \mathrm{~m}\)
- C \(\frac{37}{2} \mathrm{~m}\)
- D \(\frac{25}{2} \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(\frac{45}{4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Suppose that both the balls meet at a distance \(h\) from ball-l after time \(t\) from the start as shown in the figure, For downward motion of ball-1, from second equation of the motion,…
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