COMEDK · Physics · 4. Motion In Two Dimensions
A body is projected vertically upwards. The times corresponding to height \(h\), while ascending and while descending are \(t_{1}\) and \(t_{2}\), respectively. Then, the velocity of projection is ( \(g\) is acceleration due to gravity)
- A \(\frac{g \sqrt{t_{1} t_{2}}}{2}\)
- B \(\frac{g\left(t_{1}+t_{2}\right)}{2}\)
- C \(g \sqrt{t_{1} t_{2}}\)
- D \(\frac{g t_{1} t_{2}}{t_{1}+t_{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{g\left(t_{1}+t_{2}\right)}{2}\)
Step-by-step Solution
Detailed explanation
Let \(u\) be the initial velocity of vertical projection and \(t\) be the time taken by the body to reach at height \(h\) from the ground. \(\therefore\) Using equation, \(h=u t+\frac{1}{2} a t^{2}\) Here, \(u=u, a=-g\) \(\therefore \quad h=u t-\frac{1}{2} g t^{2}\)…
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