Question:
A ball projected vertically upwards with velocity 'v' passes through a point P in its upward journey in a time of 'x' seconds. Then, the time in which the ball again passes through the same point P is
A ball projected vertically upwards with velocity 'v' passes through a point P in its upward journey in a time of 'x' seconds. Then, the time in which the ball again passes through the same point P is
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In vertical projectile motion, time to ascend equals time to descend for the same height.
Updated On: Jun 4, 2025
- $\frac{v}{2g}$
- $\frac{2v}{g} - x$
- $\frac{v}{2g} - x$
- $2\left(\frac{v}{g} - x\right)$
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The Correct Option is D
Solution and Explanation
The time to reach point P in upward journey is $x$. The total time to reach maximum height is $\frac{v}{g}$. The ball takes equal time to come down from maximum height to point P. Hence, total time after point P is twice the time from $x$ to $\frac{v}{g}$. So, time taken to pass through point P again is: \[ 2\left(\frac{v}{g} - x\right) \]
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