A ball is projected upwards from a height h above the surfa…

Physics Motion in a Straight Line Mix Single Correct MCQ

Published on: September 12, 2026

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

\(\frac{v}{g} + \frac{2hg}{\sqrt{3}}\)

\(\frac{v}{g} \left[ 1 - \sqrt{1 + \frac{2gh}{v^2}} \right]\)

\(\frac{v}{g} \left[ 1 + \sqrt{1 + \frac{2gh}{v^{2}}} \right]\)

\(\frac{v}{g} \left[ 1 + \sqrt{v^{2} + \frac{2gh}{v}} \right]\)

Text Solution

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The correct answer is:

Since direction of v is opposite to the direction of g and h so from equation of motion

\(h = -vt + \frac{1}{2}gt^{2}\)

\(\Rightarrow gt^{2} - 2vt + 2h = 0\)

\(\Rightarrow t = \frac{-2v \pm \sqrt{4v^{2} + 8gh}}{2g}\)

\(\Rightarrow t = \frac{v}{g} \left[ 1 + \sqrt{1 + \frac{2gh}{v^{2}}} \right]\)

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