A particle is projected at an angle of 60° above the horizontal with a speed of 10 m s- 1 . After some time the direction of its velocity makes an angle of 30° above the horizontal. The speed of the particle at this instant is

Question

A particle is projected at an angle of 60° above the horizontal with a speed of 10 m s- 1 . After some time the direction of its velocity makes an angle of 30° above the horizontal. The speed of the particle at this instant is (A) (5/√3)ms- 1 (B) 5√3 m s- 1 (C) 5 m s- 1 (D) (10/√3)ms- 1

Tardigrade Admin · Accepted Answer

Let v be the velocity of point where it make angle text30 textÎ¿ with horizontal thus , horizontal velocity remains uncharged, so <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-fk73zuapsbq9d7ml.jpg" /> v cos 30 textÎ¿ = 10 cos 60 textÎ¿ v (√3/2) = ( text10/2) v=( text10/√3)ms- 1

Q. A particle is projected at an angle of $6 0^{\circ}$ above the horizontal with a speed of $10 m s^{- 1}$ . After some time the direction of its velocity makes an angle of $3 0^{\circ}$ above the horizontal. The speed of the particle at this instant is

$\frac{5}{3} m s^{- 1}$

$53 m s^{- 1}$

$5 m s^{- 1}$

$\frac{10}{3} m s^{- 1}$

Let $v$ be the velocity of point where it make angle $30^{ο}$ with
horizontal
thus , horizontal velocity remains uncharged, so

$v cos 3 0^{ο} = 10 cos 6 0^{ο}$
$v \frac{3}{2} = \frac{10}{2}$
$v = \frac{10}{3} m s^{- 1}$

Q. A particle is projected at an angle of 60∘ above the horizontal with a speed of 10ms−1 . After some time the direction of its velocity makes an angle of 30∘ above the horizontal. The speed of the particle at this instant is

3​5​ms−1

53​ms−1

5ms−1

3​10​ms−1