COMEDK · Physics · 3. Motion In One Dimension
A body is projected vertically upwards. The times corresponding to height \(h\), while ascending and descending are \(t_{1}\) and \(t_{2}\), respectively. Then, the velocity of projection is ( \(g\) is acceleration due to gravity)
- A \(g \sqrt{t_{1} t_{2}}\)
- B \(\frac{g t_{1} t_{2}}{t_{1}+t_{2}}\)
- C \(\frac{g \sqrt{t_{1} t_{2}}}{2}\)
- D \(\frac{g\left(t_{1}+t_{2}\right)}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{g\left(t_{1}+t_{2}\right)}{2}\)
Step-by-step Solution
Detailed explanation
Let \(u\) be the initial vertical velocity of the body by which it is projected vertically upward. If \(t\) be the time taken by the body to reach a height \(h\) from the point of projection, then using equation of vertical motion,…
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