#$&*
Phy 201
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_08.2_labelMessages **
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
How high does it rise and how long does it take to get to its highest point? answer/question/discussion:
v0=15 m/s, vf=0 m/s, ds=12 m, a=-10 m/s^2
dt=dv/a=(-15 m/s)/(-10 m/s^2)=1.5
ds=vAve*dt=(7.5 m/s)*1.5 s=11.25 m #$&*
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 = +15 m/s`ds = -12 ma = -10 m/s^2
vf^2 = v0^2 + 2 a `ds
vf = sqrt( v0^2 + 2 a `ds)= sqrt( (15 m/s)^2 + 2 * (-10 m/s^2) * (-12 m) )= 21.6 m/s We know that this value can not be positive so we have to make it negative.
dt=ds/vAve
vAve=(15 m/s+-21.6 m/s)/2=-3.3 m/s
dt=ds/vAve=(-12 m)/(3.3 m/s)=3.6 s
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At what clock time(s) will the speed of the ball be 5 meters / second? answer/question/discussion:
Because the speed is 5 m/s the velocity can either be positive or negative
dt=(vf-v0)/a=(5 m/s-15 m/s)/-10 m/s=1 second
dt=-5 m/s-15 m/s/-10 m/s/=2 second #$&*
At what clock time(s) will the ball be 20 meters above the ground?
To achieve a 20 meter height the vertical position must change by 8 meters because we started at 12 m and we increase to 8 m.
we use vf^2=v0^2+2a*ds to get the final velocity
when we solve we get 8.1 m/s
dt=dv/a=(-6.9 m/s)/-10 m/s^2=.69
If the velocity is -8 m/s we get 2.31 s
How high will it be at the end of the sixth second?answer/question/discussion:
After 6 m/s the ball has a velocity of -45 m/s
Average velocity is -15 m/s and the height would be -15 m/s*6 s=-90 m
The is lower than the ground???
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@&
That's right.
Maybe down a well?
*@
*#&!
&#This looks good. See my notes. Let me know if you have any questions. &#