PHY 231
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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
s0 = 12m
a = -10m/s
'dt = 1.5
'dv = -15m/s
vf = 0m/s
v_avg = 7.5m/s
'ds = 11.25m
It will rise for 1.5 seconds.
It will travel 11.25m above the origin. (A total height of 23.25m above the ground)
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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: ->->->->->->->->->->->-> :
The Initial speed going up is reversed when the ball passes the same point coming down. So the downward speed at height 12m is -15m/s. From here we infer:
'ds = -12m
`ds = v0 `dt + .5 a `dt^2
-12m = (15m/s)*'dt + (-5m/s)*'dt^2
0 = (-5m/s)*'dt^2 + (15m/s)*'dt + 12m
Using the quadratic formula we get the values of 'dt to be:
-0.66s, 3.66s
We choose the positive because we know the time is after the ball has been thrown and thus positive.
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
This is obviously 1 second. However:
vf = 5m/s
solve for 'dt
'dv = -10m/s
'dv/a = 1s
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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-12m = 8m
0 = (-5m/s)*'dt^2 + (15m/s)*'dt + 8m
Using the quadratic formula we get the values of 'dt to be:
0.7s, 2.3s
Want 'ds
'dt = 6
'dv = -60m/s
vf = -45m/s
v_avg = -15m/s
'ds = -90
add the s displacement you have -78m (assuming the ball can burrow into the ground)
or that there's a hole
#$&*
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10 minutes
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&#Your work looks very good. Let me know if you have any questions. &#