Your 'cq_1_8.2' 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 approxomation 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?
o To get to the highest point it takes 1.5 s.
T = -v0/ a
T = -15 m/s / -10 m/s^2
T = 1.5s
The ball rises 22.5 m from the person’s hand.
(15 m/s) (1.5s) = 22.5 m
The ball doesn't travel at 15 m/s for the entire 1.5 seconds. This is its speed only at the first instant. It ends up with speed 0.
To determine this we didn’t add the height of 12 meters above the ground. If we added this we would get that the ball rose 34.5 meters above the ground. The question doesn’t state if it is looking for the height from the toss or from the ground so here is both instances!!
Very good. It would be sufficient to specify either, but it's best to specify both, as you have done. See my previous note, though.
• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
o After the initial toss the ball will strike the ground at 3 seconds. This is determined by taking the 1.5 s at the highest point and adding 1.5 s. We do this because 1.5s is half the time it takes the ball to go up and fall back to its original position. Therefore, it takes the same amount of time to reach the ground from the max point.
The ground is 12 m below the initial point, so the times won't be the same.
o The ball is going a velocity of -35 m/s when it hits the ground.
o 15 m/s + (-10 m/s^2)(5.00 s) = -35 m/s
correct for the t = 5.00 second instant, but the ball isn't yet at the ground.
• At what clock time(s) will the speed of the ball be 5 meters / second?
o The speed of the ball will be 5 m/s at t = 1s
o I worked backwards from the equation:
o 15 m/s + (-10 m/s/s)(1.00s) = 5 m/s
Very good, but the speed is also 5 m/s when the velocity is -5 m/s. Speed is the absolute value, or magnitude, of the velocity.
• At what clock time(s) will the ball be 20 meters above the ground?
o With a process of trial and error we see that the ball will be 20 meters above ground at 0.5 s. We calculated this by observing that the ball is 34.5 m above the ground at 1.5s and 15 m above the ground at 1s. We found that the ball is 20 above the ground at this point because we took 15 m/s * .5 s and obtained 7.5 m. We then had to add the height given of 12 m above the ground. By adding the two together we rounded to get 20 meters above the ground at t = 0.5 s.
Your basic idea would be correct if the ball maintained its 15 m/s velocity while rising. However, this is not the case.
• How high will it be at the end of the sixth second?
o 15 m/s + (-10 m/s/s) (6.00s) = -45 m/s (6.00s) = -270 m above ground
-45 m/s is the final velocity.
The ball does not move for 6.00 s at its final velocity.
If you average initial and final velocities to get the average velocity you will get the correct result.
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1 hour
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Again I'm not sure about the last question.
You did a lot of good work and good thinking on this question. You did miss a few details.
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