Question

A gun is placed on a cliff of height 60 m above ground level, and fires a projectile at an angle of 50 degrees. The muzzle velocity of the projectile is 139 m/s.

What will be the magnitude of the total velocity when it hits the level ground (y=0m).

Round your answer to the nearest tenth of m/s.

Answer 1

Height of the cliff H = 60 m

Angle at which projectile is fired = 50°

Initial velocity of the projectile is u = 139 m/s

We can resolve the initial velocity into horizontal and vertical components.

Horizontal component of the velocity is u_x = u cos

= 139 * cos50°

= 139 * 0.642

= 89.23 m/s

Vertical component of the velocity is u_y = u sin

= 139 * sin50°

= 106.48 m/s

In the horizontal direction no force acts on the projectile . So horizontal component of the velocity remains constant at any instant of time and is same when the projectile touches the ground

In the vertical direction , gravity acts and so the body will have acceleration due to gravity

So the final vertical velocity, v_y² = u_y² + 2 g h

= ( 106.48 )² + 2 * 9.81* 60

= 12515.19

v_y = 12515.19

v_y = 111.87 m/s

This is the vertical component of the velocity when the projectile touches the ground

So total velocity of the projectile when it touches the ground is

V = u_x² + v_y²

= ( 89.23² + 111.87² )

=  7961.99 + 12515.19

=   20477.18

= 143.09

= 143.1 m/s

So the velocity of the projectile when it hits the ground is about 143 m/s