A projectile is thrown upward at an angle 60° with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is 45° with the horizontal. Find the initial speed of the particle.

Given:

• Angle of projection = 60°
• Speed when direction becomes 45° = 20 m/s

Horizontal velocity remains constant:

u_x = u cos 60° = u/2

At the instant when direction is 45°:

v_x = v_y

Given resultant speed = 20 m/s

v_x = v_y = 20/√2 = 10√2

Since horizontal velocity remains constant:

u/2 = 10√2

u = 20√2 m/s

Answer: 20√2 m/s