Back to Directory
NEET PHYSICSMedium

A ball is projected with a velocity, 10 ms−110 \text{ ms}^{-1}, at an angle of 60∘60^\circ with the vertical direction. Its speed at the highest point of its trajectory will be:

A

Zero

B

53 ms−15\sqrt{3} \text{ ms}^{-1}

C

5 ms−15 \text{ ms}^{-1}

D

10 ms−110 \text{ ms}^{-1}

Step-by-Step Solution

The angle with the vertical is 60∘60^\circ, so the angle with the horizontal is θ=90∘−60∘=30∘\theta = 90^\circ - 60^\circ = 30^\circ. At the highest point, the vertical component of velocity is zero, and the speed is equal to the horizontal component: v=ucos⁡(θ)=10cos⁡(30∘)=10×32=53 ms−1v = u \cos(\theta) = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ ms}^{-1}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started