Question

A particle is projected from the horizontal at an inclination of ${60}^o$ with an initial velocity $20m/s$. Find the time at which the energy becomes three-fourth kinetic and one-fourth potential energy. (in seconds).

• A. $\sqrt{2}+\sqrt{3}$
• B. $\sqrt{2}-\sqrt{3}$
• C. $\sqrt{2}+\sqrt{5}$
• D. $\sqrt{7}+\sqrt{3}$

Answer 1

Answer: (A), (B)

$\sqrt{2}+\sqrt{3}$
$\sqrt{2}-\sqrt{3}$

Initially kinetic energy of the body is $\frac{1}{2}m{v}^2$ $=200m$

If potential energy is to be one-fourth of total energy then $P.E.=50m$

Potential energy is given by $mgh$ $=>mgh=50m$ taking $g=10$ we get $h=5m$

So the body will be at this height for 2 times once going upwards and then coming downwards.

Vertical component of velocity is $vSin\theta =10\sqrt{3}$. Acceleration is $-10m/s^2$, displacement is$5m$

So using $S=u.t+\frac{1}{2}a{t}^2$ we get $5=10\sqrt{3}t-5t^2$

So $t=\sqrt{3}+\sqrt{2}and\sqrt{2}-\sqrt{3}$

So option A and B both arr correct.