Skill Set 03.2

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1. A ball rolls off the edge of a cliff. The instant it leaves the edge, it has an initial horizontal velocity of 17.3 m/s, and it strikes the ground after 1.56 seconds. How high is the cliff?

Using s = ut + 1/2at²:

s = 0 + 1/2(9.81)(1.56)² = 11.936 m

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2. A ball rolls off the edge of a cliff. The instant it leaves the edge, it has an initial horizontal velocity of 17.3 m/s, and it strikes the ground after 1.56 seconds. How far from the base of the cliff does the ball land?

We will again use s = ut + 1/2at², but since horizontal acceleration = 0, the equation becomes s = ut.

s = ut = (17.3)(1.56) = 26.988 m

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3. Question

Solution goes here

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4. Question

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5. Question

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