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 trikes the ground is

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

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{v}{g}+\frac{2hg}{\sqrt{2}}$
B $\frac{v}{g}[1-\sqrt{1+\frac{2h}{g}}]$
C $\frac{v}{g}[1+\sqrt{1+\frac{2gh}{v^2}}]$
D $\frac{v}{g}[1+\sqrt{v^2+\frac{2g}{h}}]$

The time taken is given as,

$v=u+gt$

$v=0+gt$

$\frac{v}{g}=t$

The height is given as,

$0-v^2=-2gh$

$\frac{v^2}{2g}=h$

The total height attained is given as,

$s=\frac{v^2}{2g}+h$

$s=ut+\frac{1}{2}g{t}^2$

$\frac{v^2}{2g}+h=\frac{1}{2}g{t_1}^2$

$\sqrt{\frac{2}{g}\left(\frac{v^2}{2g}+h\right)}=t_1$

The total time taken is given as,

$t+t_1=\frac{v}{g}+\sqrt{\frac{2}{g}\left(\frac{v^2}{2g}+h\right)}$

$t+t_1=\frac{v}{g}\left[1+\sqrt{\left(v^2+\frac{2g}{h}\right)}\right]$

Thus, the time at which ball strikes the ground is $\frac{v}{g}\left[1+\sqrt{\left(v^2+\frac{2g}{h}\right)}\right]$.