Question

A shell is fired at an angle of elevation of 300 with a velocity of 500 m s–1. Calculate the vertical and horizontal components of the velocity, the maximum height that the shell reaches, and its range.

Answer 1

Question : A shell is fired at an angle of elevation of 30° with a velocity of 500 m/s . Calculate the vertical and horizontal components of the velocity, the maximum height that the shell reaches, and its range.

Answer :

Initial velocity = 500m/s

Angle of projection = 30°

Horizontal component of velocity

= V cosθ

= 500 × cos30

= 500 × (√3 /2)

= 250√3 m/s

Vertical component of velocity

= V sinθ

= 500 × sin30

= 500 × 1/2

= 250 m/s

Maximum Height attained by a projectile is given by,

H = u² sin²θ / 2g

H = 500² * (sin30)² / 2(10)

H = 250000 * 1/4 * 1/20

H = 3125 m.

Range of the projectile is related to Maximum height as,

tanθ = 4H / R

tan30 = 4(3125) / R

R = 12500 / ( 1/√3)

R = 12500 √3 m.

Therefore,

• Horizontal component of velocity = 250√3 m/s
• Vertical component of velocity = 250 m/s
• Maximum Height attained by shell = 3125 m.
• Range of the shell = 12500√3 m