Question

A particle is projected with velocity of 40√2 m/s making anangle of 45° with the horizontal, the magnitude of velocityof the particle at t = 1 s after projection, will be​

Answer 1

Given :

▪ Initial velocity = 40√2 m/s

▪ Angle of projection = 45°

To Find :

▪ Velocity of particle after 1s.

Concept :

☞ In projectile motion x - component of velocity remains constant throughout the motion but y - component of velocity changes with time as acceleration due to gravity continuously acts in downward direction.

☞ In projectile motion, velocity of projectile after time t is given by

⭐ V = √(Vx² + Vy²)

where,

Vx = ucosΦ

Vy = usinΦ - gt

(Φ denotes angle of projection)

Calculation :

✴ x-component of velocity :

→ Vx = ucosΦ

→ Vx = 40√2cos45°

→ Vx = 40√2×1/√2

→ Vx = 40m/s

✴ y-component of velocity :

→ Vy = usinΦ - gt

→ Vy = 40√2sin45° - (10×1)

→ Vy = 40√2×1/√2 - 10

→ Vy = 40 - 10

→ Vy = 30m/s

✴ Final velocity :

→ V = √(Vx² + Vy²)

→ V = √(40² + 30²)

→ V = √(1600 + 900)

→ V = √(2500)

→ V = 50m/s