Question

A particle is thrown vertically upwards. Its velocity at one-fourth of the maximum height is 20 m$s^{-1}$.Then, the maximum height attained by it is

[take g = 10 m$s^{-2}$]

• A. 16.67 m
• B. 10.67 m
• C. 26.67 m
• D. 25 m

Answer 1

The correct option is C 26.67 m

Let maximum height is $h$ and the initial velocity is $u$.

At maximum height final velocity $v=0$,

Using $v^2=u^2+2as$ $\Rightarrow 0=u^2-2gh$ $\Rightarrow u^2=2gh$ ......1

Now at height $\frac{h}{4}$ , velocity $v=20m{s}^{-1}$,

Using $v^2=u^2+2as$ $\Rightarrow {20}^2=u^2-2g\frac{h}{4}$

$\Rightarrow {20}^2=2gh-\frac{gh}{2}$ (Using equation 1)

$\Rightarrow h=\frac{80}{3}m$