A ball is projected horizontally from a 20 m high cliff with speed 15 m/s. Assuming g = 9.8 m/s^2 and no air resistance, approximately how far from the base will it land?

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Multiple Choice

A ball is projected horizontally from a 20 m high cliff with speed 15 m/s. Assuming g = 9.8 m/s^2 and no air resistance, approximately how far from the base will it land?

Explanation:
The key idea is that horizontal and vertical motions are independent. The ball’s horizontal speed stays constant at 15 m/s (no air resistance), while it falls under gravity from a height of 20 m. First find how long it takes to hit the ground using the vertical motion: s = (1/2) g t^2, so t = sqrt(2h/g) = sqrt(40/9.8) ≈ 2.02 s. During that time, it travels horizontally a distance d = v0 t = 15 m/s × 2.02 s ≈ 30.3 m. So it lands about 30 meters from the base.

The key idea is that horizontal and vertical motions are independent. The ball’s horizontal speed stays constant at 15 m/s (no air resistance), while it falls under gravity from a height of 20 m.

First find how long it takes to hit the ground using the vertical motion: s = (1/2) g t^2, so t = sqrt(2h/g) = sqrt(40/9.8) ≈ 2.02 s.

During that time, it travels horizontally a distance d = v0 t = 15 m/s × 2.02 s ≈ 30.3 m. So it lands about 30 meters from the base.

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