Question

A ball is thrown from the edge of a 40.0 m high cliff with a speed of 20.0 m/s at an angle of 30.0° below horizontal. What is the speed of the ball when it hits the ground below the cliff?

Answer 1

Velocity is 34.42 m/s at an angle of 56.91° below horizontal

Speed is 34.42 m/s

Step-by-step explanation:

Velocity = 20.0 m/s at an angle of 30° below horizontal

Vertical velocity = 20 sin 20 = 6.84 m/s downward.

Horizontal velocity = 20 cos 20 = 18.79 m/s towards right.

Let us consider the vertical motion of ball we have equation of motion

v² = u² + 2as

We need to find v, u = 6.84 m/s, a = 9.81 m/s² and s = 40 m

Substituting

v² = 6.84² + 2 x 9.81 x 40 = 831.59

v = 28.84 m/s

So on reaching ground velocity of ball is

Vertical velocity = 28.84 m/s downward.

Horizontal velocity = 18.79 m/s towards right.

Velocity

`v=√(28.84^2+18.79^2)=34.42m/s`

`tan\theta =(28.84)/(18.79)\\\\\theta =56.91^0`

So velocity is 34.42 m/s at an angle of 56.91° below horizontal

Speed is 34.42 m/s