Question

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

• A. $\frac{v}{g}\sqrt{1-\frac{2hg}{v^2}}$
• B. $\frac{v}{g}\sqrt{1+\frac{2hg}{v^2}}$
• C. $\sqrt{1+\frac{2hg}{v^2}}$
• D. $\frac{v}{g}[1+\sqrt{1+\frac{2hg}{v^2}}]$

Answer 1

The correct option is D $\frac{v}{g}[1+\sqrt{1+\frac{2hg}{v^2}}]$

Using Kinematics equation,
$-h=vt-\frac{1}{2}g{t}^2$

$g{t}^2-2vt-2h=0$

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

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