TS EAMCET · Physics · Motion In Two Dimensions
A body is projected vertically upwards at time \(t=0\) and it is seen at a height \(H\) at time \(t_1\) and \(t_2\) second during its flight. The maximum height attained is ( \(g\) is acceleration due to gravity)
- A \(\frac{g\left(t_2-t_1\right)^2}{8}\)
- B \(\frac{g\left(t_1+t_2\right)^2}{4}\)
- C \(\frac{g\left(t_1+t_2\right)^2}{8}\)
- D \(\frac{g\left(t_2-t_1\right)^2}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{g\left(t_2-t_1\right)^2}{8}\)
Step-by-step Solution
Detailed explanation
Let time taken by the body to fal from point \(C\) to \(B\) is \(t^{\prime}\). Then \(t_1+2 t^{\prime}=t_2\) \(t^{\prime}=\left(\frac{t_2-t_1}{2}\right) \ldots(\mathrm{i})\) Total time taken to reach point \(C\)…
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