A ball is projected with a velocity, 10 ms -1, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be :

Question

A ball is projected with a velocity, 10 ms -1, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be : (A) 10 ms -1 (B) Zero (C) 5 √3 ms -1 (D) 5 ms -1

Tardigrade Admin · Accepted Answer

At highest point only horizontal component of velocity remains ⇒ ux=u cos θ <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d742654680fcc033b3ec13aee2c9a581-.png" /> u x = u cos θ =10 cos 30° =5 √3 ms -1

Q. A ball is projected with a velocity, $10 m s^{- 1}$ , at an angle of $6 0^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be :

$10 m s^{- 1}$

Zero

$53 m s^{- 1}$

$5 m s^{- 1}$

At highest point only horizontal component of velocity remains
$\Rightarrow u_{x} = u cos θ$

$u_{x} = u cos θ = 10 cos 3 0^{\circ}$
$= 53 m s^{- 1}$

Q. A ball is projected with a velocity, 10ms−1, at an angle of 60∘ with the vertical direction. Its speed at the highest point of its trajectory will be :

10ms−1

Zero

53​ms−1

5ms−1

At highest point only horizontal component of velocity remains ⇒ux​=ucosθ ux​=ucosθ=10cos30∘ =53​ms−1

Q. A ball is projected with a velocity, $10 m s^{- 1}$ , at an angle of $6 0^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be :

$10 m s^{- 1}$

$53 m s^{- 1}$

$5 m s^{- 1}$

At highest point only horizontal component of velocity remains
$\Rightarrow u_{x} = u cos θ$

$u_{x} = u cos θ = 10 cos 3 0^{\circ}$
$= 53 m s^{- 1}$