#$&*
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.
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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: ->->->->->->->->->->->-> :
It is at it's highest point when it changes direction from moving upward to falling downward. This means that it is at its highest point when v = 0 m/s
vf^2 = v0^2 + 2 * a * 'ds 0^2 m/s = 15^2 m/s + 2 * -10 m/s^2 * 'ds
'ds = (0^2 m/s - 15^2 m/s) / -20 m/s^2 = -225 m/s / -20 m/s^2 = 11.25 meters
This is not it's highest point though because it started out at 12 meters, so 11.25 meters + 12 meters = 23.25 meters
The 'dt is found using 11.25 meters though since 23.25 meters adds on the ball's initial position.
'ds = (vf + v0) / 2 * 'dt 'dt = (vf + v0) / 2 * 'ds
'dt = (0 m/s + 15 m/s) / 2 * 11.25 meters = 0.667 seconds.
This shows that it takes 0.667 for the ball to reach the height of 23.25 meters (11.25 meters added on to the 12 meters).
I must have missed something because this doesn't seem right. It should take at least 1.5 seconds to reach a velocity of 0 seconds.
Good thinking. There is an error and you did well to spot the fact.
You multiplied (or more likely divided) an average velocity by a displacement. If you divided, then you got it upside down; easy enough to fix.
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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 'ds = -12 meters, vf^2 = 15^2 m/s + 2 * -10 m/s^2 * -12 meters
vf ^2 = 465, sqrt(vf^2) = sqrt(465) = +-21.56 m/s In this case it is - 21.56 m/s because it is heading downward.
-21.56 m/s = 15 m/s + -10 m/s^2 * 'dt 'dt = -21.56 - 15 / -10
'dt = 3.656 seconds
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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 5 m/s = 15 m/s * -10 m/s^2 * 'dt
'dt = 5 m/s - 15 m/s / -10 m/s^2 1 second
This makes since because the v0 is 15 m/s and the acceleration is -10 m/s^2. But it is only the first time the ball is going 5 m/s.
After the ball reaches it's max height (at 1.5 seconds) it will begin to descend. It reaches 5 m/s again at 2 seconds. It will travel 10 m/s due to it's acceleration and after another .5 seconds it will be at 5 m/s.
So the ball is at 5 m/s at 1 second and 2 seconds.
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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: ->->->->->->->->->->->-> :
vf^2 = v0^2 + 2 * a * 'ds vf^2 = 225 m/s + 2 * -10 m/s^s * 8 meters
vf^2 = 65 m/s sqrt(vf^2) = sqrt(65 m/s) = 8.06 m/s (the value is positive because it is still moving upward. 8.06 m/s = 15 m/s * -10 m/s^2 * 'dt, 'dt = .694 seconds
Once the ball is at it's max height of 23.25 it will fall back down to 20 meters above the ground so 'ds = -3.25 meters.
vf ^ 2 = 225 m/s + 2 * -10 m/s^2 * -3.25 meters vf = -17.02 m/s
vf = v0 + a * 'dt 'dt = -17.02 m/s - 15 m/s = -10 m/s^2 = 3.202 seconds
The ball is at 20 meters above the ground at 0.694 seconds and 3.202 seconds.
'ds = v0 * 'dt + 0.5 * a * 'dt^2 'ds = 15 m/s * 6 s + 0.5 * -10 m/s^2 * 36 s
'ds = -90 meters, so the height is 78 meters below ground.
I'm not sure how the acceleration and velocity would change after the ball would bounce so im implying that it burrows into the ground.
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20 min
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&#This looks good. See my notes. Let me know if you have any questions. &#