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Phy 201
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_08.2_labelMessages **
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: ->->->->->->->->->->->-> :
v0 = 15m/s
a = -10m/s^2
vf = 0
‘dt = (0 - 15m/s)/-10m/s^2 = 1.5s
‘ds = 7.5m/s * 1.5s +12m = 23.25m
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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: ->->->->->->->->->->->-> :
vf = ?
v0 = 0m/s
a = -10m/s^2
‘ds = 23.25m
Vf^2 = 0 + 2*-10m/s^2 * 23.25m
Vf^2 = -465m^2/s^2
Vf = -21.56m/s
It is going 21.56m/s when it hits the ground
‘dt = (-21.56m/s - 15m/s)/-10m/s^2 = 3.66s
It took 3.66s from the initial throw to reach the ground.
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Good.
This could also have been solve in one step, using
v0 = 15 m/s
`ds = -12 m
a = -9.8 m/s^2.
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a* ‘dt
5m/s = 15m/s + -10m/s^2 * ‘dt
-10m/s = -10m/s^2 * ‘dt
1s = ‘dt
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Speed is also 5 m/s when velocity is -5 m/s.
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• At what clock time(s) will the ball be 20 meters above the ground?
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
‘ds = 20m
A = -10m/s^2
V0 = 15m/s
‘dt = ?
20m = 15m/s * ‘dt + 0.5 * -10m/s^2 * ‘dt^2
20m = 15m/s * ‘dt - 5m/s^2*’dt^2
-5m/s^2 * ‘dt^2 + 15m/s * ‘dt - 20m = 0
hmm
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This is a quadratic equation of the form a t^2 + b t + c. It can be solved using the quadratic formula.
You'll get two times, one corresponding to the ball when rising, the other while falling back down.
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25 mins
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&#Good responses. See my notes and let me know if you have questions. &#