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Solution


Q.

A ball is projected from point A with velocity 20ms-1 at an angle 60° to the horizontal direction. At the highest point B of the path (shown in figure), the velocity vms-1 of the ball will be:           [2023]
1 20  
2 103  
3 Zero  
4 10  

Ans.

(4)

Given, u=20m/s; θ=60°; v=?

At highest point, the object doesn't have vertical velocity, it has only horizontal velocity and moves in the horizontal direction.

So, θ'=0°

At θ=60°, horizontal velocity =vcosθ'

During motion, its horizontal component remains constant

i.e., ucosθ=vcosθ'

or 20cos60°=vcos0°

v=10m/s

The velocity of the projectile at the highest point is 10 m/s.