A particle is projected with velocity 2 √g h and at an angle 60° to the horizontal so that it just clears two walls of equal height h which are at a distance 2 h from each other. The time taken by the particle to travel between these two walls is

Question

A particle is projected with velocity 2 √g h and at an angle 60° to the horizontal so that it just clears two walls of equal height h which are at a distance 2 h from each other. The time taken by the particle to travel between these two walls is (A) 2 √(2 h/g) (B) √(h/2 g) (C) 2 √(h/g) (D) √(h/g)

Tardigrade Admin · Accepted Answer

Given, velocity of particle =2 √g h Angle (θ)=60° Distance covered by particle =2 h Now, 2 h=2 √g h cos 60° t or, √h=√g × (1/2) t or, t=2 √(h/g)

Q. A particle is projected with velocity $2 g h$ and at an angle $6 0^{\circ}$ to the horizontal so that it just clears two walls of equal height $h$ which are at a distance $2 h$ from each other. The time taken by the particle to travel between these two walls is

$2 \frac{2 h}{g}$

$\frac{h}{2 g}$

$2 \frac{h}{g}$

$\frac{h}{g}$

Given, velocity of particle $= 2 g h$
Angle $(θ) = 6 0^{\circ}$
Distance covered by particle $= 2 h$
Now, $2 h = 2 g h cos 6 0^{\circ} t$
or, $h = g \times \frac{1}{2} t$
or, $t = 2 \frac{h}{g}$

Q. A particle is projected with velocity 2gh​ and at an angle 60∘ to the horizontal so that it just clears two walls of equal height h which are at a distance 2h from each other. The time taken by the particle to travel between these two walls is

2g2h​​

2gh​​