COMEDK · Physics · 3. Motion In One Dimension
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 will be (take, \(g\) as acceleration due to gravity)
- A \(g \dfrac{t_1 t_2}{\left(t_1+t_2\right)}\)
- B \(\dfrac{g \sqrt{t_1 t_2}}{2}\)
- C \(g \sqrt{t_1 t_2}\)
- D \(\dfrac{g\left(t_1+t_2\right)}{2}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{g\left(t_1+t_2\right)}{2}\)
Step-by-step Solution
Detailed explanation
Let \(u\) be the velocity of projection. The equation of motion for a body at height \(h\) is given by \(h = ut - \dfrac{1}{2}gt^2\). Rearranging the terms, we get a quadratic equation in \(t\): \(\dfrac{1}{2}gt^2 - ut + h = 0\). The roots of this equation are \(t_1\) and…
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