A ball is thrown vertically upwards with a speed v from height h meter above the ground….

Q: A ball is thrown vertically upwards with a speed v from height h meter above the ground. The time taken for the ball to hit ground is

(a) \displaystyle \frac{v}{g}\sqrt{1-\frac{2hg}{v^2}}

(b) \displaystyle \frac{v}{g}\sqrt{1+\frac{2hg}{v^2}}

(c) \displaystyle \sqrt{1-\frac{2hg}{v^2}}

(d) \displaystyle  \frac{v}{g}[1 + \sqrt{1+\frac{2hg}{v^2}}]

Ans: (d)

Sol: \displaystyle -h = vt - \frac{1}{2}gt^2

\displaystyle  -2h = 2vt - gt^2

\displaystyle  gt^2 - 2vt - 2h = 0

\displaystyle  t = \frac{2v \pm \sqrt{4v^2 - 4g(-2h)}}{2g}

\displaystyle  t = \frac{v}{g}+ \frac{1}{2g}\sqrt{4v^2 + 8gh}

\displaystyle t =  \frac{v}{g} + \frac{v}{g}\sqrt{1+\frac{2hg}{v^2}}

\displaystyle t =  \frac{v}{g}[1 + \sqrt{1+\frac{2hg}{v^2}}]

Author: Rajesh Jha

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