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Phy 121
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: ->->->->->->->->->->->-> :
Given: v0=15m/s, vf=0m/s(for highest point), a=-10m/s
Find ‘dt
vf=v0 + a ‘dt
(vf-v0)/a=’dt
(0m/s-15m/s)/ -10m/s^s=1.5s
It takes the ball 1.5s to reach it’s highest point.
The ball’s highest point is at ‘dt=1.5s and vf=0m/s:
‘ds=((v0+vf))/2*’dt
‘ds=((15m/s+0m/s)/2)*1.5s=11.25m
The ball’s highest point is 11.25m + 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: ->->->->->->->->->->->-> :
When the ball hits the ground it is going:
vf^2=v0^2 + 2 a ‘ds
vf=sqrt(v0^2 + 2 a ‘ds)
vf=sqrt(0^2 + 2 * -10m/s^2 * 23.25m)=-+21.56m/s
When the ball hits the ground it’s going 21.56m/s down.
Find the time from toss to ground
1.5s from toss to highest point.
Find ‘dt from highest point to ground, then add 1.5s
‘ds=v0 ‘dt + .5 a ‘dt^2
‘dt=sqrt((‘ds-v0)/(.5*a))
‘dt=sqrt((23.25m - 0m/s)/(.5*-10m/s^2)=2.16s
Time from toss to ground is 1.5s + 2.16s= 3.66s
@&
Good.
It wasn't necessary to break this into two separate phases of motion. For example:
v_0 = 15 m/s
`ds = -12 m
a = -9.8 m/s^2
define the motion for the interval from release to striking the ground 12 m below the point of release.
This situation actually has two solutions, one with positive `dt and one with negative `dt. The former applies to this situation; the latter describes the motion of a particle which might have started from the ground a time `dt before it reached the 12 meter height, where its upward velocity would have fallen to 15 m/s.
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf^2=v0^2 + 2 a ‘dt
‘dt=(vf-v0)/a
‘dt=(5m/s - 15m/s)/(/10m/s^2)=1s, at 1s on the way up
‘dt=(5m/s-0m/s)/(10m/s^2)=.5s, at .5s after starting down from max height. Clock time of 1.5s
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Speed is also 5 m/s if the velocity is -5m/s. So there are two times when the speed is 5 m/s, once on the way up and once on the way down.
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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: ->->->->->->->->->->->-> :
Find ‘dt for 20m above ground. Which is 8m higher than starting point of 12m from ground.
‘ds=v0 ‘dt + .5 a ‘dt^2
‘dt=sqrt((‘ds-v0)/(.5*a))
‘dt=sqrt((8m - 15m/s)/(.5*-10m/s^2)=1.18s
At clock time 1.18s the ball is 20m above the ground
Find how high at end of 6 seconds:
‘ds=v0 ‘dt + .5 a (‘dt)^2
‘ds=(15m/s*6s) + (.5*-10m/s^2*(6s)^2)=-90m
We calculated earlier that the ball hit the ground after 3.66s
Not surprising that at 6s we have a negative number. Looks like we went under ground if that was possible.
#$&*
&#Your work looks good. See my notes. Let me know if you have any questions. &#