TS EAMCET · Physics · Motion In One Dimension
A stone is dropped from a height of \(100 \mathrm{~m}\), while another one is projected vertically upwards from the ground with a velocity of \(25 \mathrm{~m} / \mathrm{s}\) at the same time. The time in seconds after which they will have the same height is (acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )
- A 4
- B 5
- C 6
- D 7
Answer & Solution
Correct Answer
(A) 4
Step-by-step Solution
Detailed explanation
Let stones meet after time \(t\) seconds. Then distance travelled by first stone in \(t\) seconds \(=100-h(\mathrm{~m})\) and distance travelled by Ind stone in \(t\) second is \(h(\mathrm{~m})\). So, we have For first stone, \(\quad-(100-h)=\frac{-1}{2} g t^2\) For second…
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