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

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

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

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

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

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

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

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