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course Phy 121
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity). •How high does it rise and how long does it take to get to its highest point?
answer/question/discussion:
0m/s = 15m/s + -10m/s/s * `dt
-15m/s = -10m/s/s *`dt
(-15m/s)/(-10m/s/s) = `dt = 1.5s
`ds = (15m/s) / 2 * 1.5s
`ds = 11m
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• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
answer/question/discussion:
v0 = 0
`ds = 23m
A = 9.8m/s/s
Vf^2 = 2(9.8m/s/s)(23m)
Vf^2 = 450.8m^2/s^2
Vf = 21.2m/s
21.2m/s / 2 = 10.6m/s = vAve
`dt =23m/10.6m/s = 2.2s
2.2s + 1.5s = 3.7s
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• At what clock time(s) will the speed of the ball be 5 meters / second?
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You know v0 and a. If you know also that | vf | = 5 m/s, what then will be `dt?
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answer/question/discussion:
???????
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• At what clock time(s) will the ball be 20 meters above the ground?
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Again you know v0 and a. What `ds will put you 20 meters above the ground?
Knowing v0, a and `ds, you can then find vf and `dt.
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• How high will it be at the end of the sixth second?
answer/question/discussion:
?????
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Self-critique (if necessary):
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Self-critique rating:
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Good start. Check my notes and see what you can do with the last few questions.
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