AP EAMCET · PHYSICS · Motion In One Dimension
A ball is projected upwards from a height \(h\) above the surface of the earth with velocity \(v\). The time at which the ball strikes the ground is
- A \(\frac{v}{g}\left[1+\sqrt{\frac{2 g h}{v^2}}\right]\)
- B \(\frac{v}{g}\left[1-\sqrt{1+\frac{2 h}{g}}\right]\)
- C \(\frac{v}{g}\left[1+\sqrt{1+\frac{2 g h}{v^2}}\right]\)
- D \(\frac{v}{g}\left[1+\sqrt{v^2+\frac{2 g}{v^2}}\right]\)
Answer & Solution
Correct Answer
(C) \(\frac{v}{g}\left[1+\sqrt{1+\frac{2 g h}{v^2}}\right]\)
Step-by-step Solution
Detailed explanation
The given situation is shown in the following figure. If time taken by the ball to reach at highest point \(A\) is \(t\), then \(0=v-g t \Rightarrow t=\frac{v}{g}\) Total time taken to reach the ball from point of projection to reach at point \(B\) is given as…
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