Difference between revisions of "Skill Set 03.2"
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− | ===2. | + | ===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?=== |
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− | + | We will again use s = ut + 1/2at<sup>2</sup>, but since horizontal acceleration = 0, the equation becomes s = ut.<br><br> | |
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+ | s = ut = (17.3)(1.56) = '''26.988 m''' | ||
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===3. Question=== | ===3. Question=== | ||
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Revision as of 10:32, 12 February 2009
Main Page > IB Physics Skill Sets > Skill Set 03.2
Contents
- 1 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?
- 2 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?
- 3 3. Question
- 4 4. Question
- 5 5. Question
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
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
3. Question
Solution goes here
4. Question
Solution goes here
5. Question
Solution goes here