Question

A body projected horizontally with a velocity $v$ from a height $h$ has a range $R$. With what velocity this body has to be projected horizontally from a height $\frac{h}{2}$ to have the same range?

• A. $\sqrt{2}v$
• B. $2v$
• C. $6v$
• D. $8v$

Answer 1

Answer: (A)

$\sqrt{2}v$

In $y$ direction:
$h=ut+\frac{1}{2}g{t}^2$

So, $t=\sqrt{\frac{2h}{g}}$
In $x$ direction
$R=v\times t$

Now, In $y$ direction
$\frac{h}{2}=uT+\frac{1}{2}g{T}^2$
$T=\sqrt{\frac{h}{g}}$

So, comparing range:
$v\times \sqrt{2h/g}=v_1\times \sqrt{h/g}$
$v_1=v\sqrt{2}$