Question

a body is projected with initial velocity 10 root 3 m/sec making the angle of 30 degree with the horizontal the velocity of the particle at the highest point of the trajectory is (a) 15m/sec​

Answer 1

Given:

Initial velocity of the body = 10√3 m/s

Angle that the body makes with horizontal when projected = 30°

To find:

The velocity of the body at the highest point of its trajectory in the projectile motion.

Solution:

Let the velocity of the body at any point in its trajectory be V→ = Vx i + Vy j

Therefore V = V cosα i + V sinα j

Where α is the angle the body makes with the horizontal.

V→ = 10√3 cos 30°i + 10√3 sin 30°j

= 10√3 * √3/2 i + 10√3*1/2 j

= 15 i + 5√3 j

At the highest point the body will only have horizontal velocity:

Therefore V = 15 m/s i

Therefore the velocity of the particle at the highest point of its trajectory will be 15 m/s i.