Question

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

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

Answer 1

Answer: (C)

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

Answer 2

Since direction of v is opposite to the distance of g and h so from equation of motion h=-vt$+\frac{1}{2}gt^2\rightarrow gt^2-2ut-2h=0$ $\Rightarrow t=\frac{2v \pm \sqrt{4v^2+8gh}}{2g} \Rightarrow t=\frac{v}{g} \left[1+\sqrt{1+\frac{2gh}{v^2}}\right]$