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

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

Speed is 34.42 m/s

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

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

Speed is 34.42 m/s