Question

A body, projected vertically up with a velocity of $20m{s}^{-1}$, reaches a height of $20m$. If it is projected with a velocity of $40m{s}^{-1}$, then the maximum height reached by the body is:

• A. $20m$
• B. $10m$
• C. $80m$
• D. $40m$

Answer 1

Answer: (C)

$80m$

We know, $v^2=u^2-2gs$

$v=0$ for maximum height.

Since $g$ is constant, $s\propto u^2$.

So, $\frac{s_1}{s_2}=(\frac{u_1}{u_2}{)}^2$

$u_1=20m/s$, $u_2=40m/s$, $s_1=20m$

$s_2=(\frac{40}{20}{)}^2\times 20=4\times 20=80m$