TS EAMCET · Physics · Motion In One Dimension
A ball is thrown vertically upward from the ground at time, \(t=0 \mathrm{~s}\). It passes the top of a tower at \(t=3 \mathrm{~s}\) and \(2 \mathrm{~s}\) later it reaches and its maximum height. The height of the tower is (Acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(105 \mathrm{~m}\)
- B \(125 \mathrm{~m}\)
- C \(85 \mathrm{~m}\)
- D \(65 \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(105 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
According to question, ball reaches at top of a tower at the time, \(t_1=3 \mathrm{~s}\) and further reaches at the maximum height at time, \(t_2=2 \mathrm{~s}\). \(\therefore\) Total time taken by ball to reached the maximum height,…
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