Question

Someone stands at the edge of a cliff 15 m high. He drops a ball. How fast is the ball moving at the foot of the cliff, just before it hits the ground?
Now the person throws the ball at an angle of 43.6978° to the horizontal, at 5.0 m/s. The ball lands at the foot of the cliff after it moves in a parabolic path. How fast is the ball moving when it hits the ground at the foot of the cliff?

Answer 1

The final velocity of the ball when it falls from the 15 m high cliff is 17.15 m/s.

The final velocity of the ball when it is projected at an initial velocity of 5 m/s at 43.6978° to the horizontal is 17.5 m/s.

What is the final velocity of the ball?

The final velocity of the ball when it falls from the 15 m high cliff is calculated as follows;

v² = u² + 2gh

where;

• v is the final velocity
• u is the initial velocity = 0
• g is the acceleration due to gravity
• h is height of fall

v² = 0² + 2gh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 15)

v = 17.15 m/s

The final velocity of the ball when it is projected at an initial velocity of 5 m/s at 43.6978° to the horizontal.

v² = (5 sin43.6978)² + 2(9.8)(15)

v² =  305.93

v = √305.93

v = 17.5 m/s

Learn more about final velocity here: brainly.com/question/25905661

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