ender7x77 Posted February 27, 2007 Posted February 27, 2007 A ball is thrown vertically upwards. It rises, falls back down, strikes the floor and bounces back up hwere it is caught at the same position of that it was originally released. Assume that no energy is lost when the ball hits the floor. If anyone can help the aforementioned is to be graphed, but by no means am I asking you to do it for me. The problem is that the graph has to have a scale. Now, I figure a velocity time graph would start above the time axis, to come in contact with the time axis, to go below the time axis, to return back to the time axis, and then would end above the time axis. I am also given the initial speed (10m/s), where it is released (2m above the ground), and the acceleration is assumed to be 10m/s^2 [down]. If anyone can point in the right direction in order to make a scale for this graph...it would be greatly appreciated. thanks.
swansont Posted February 27, 2007 Posted February 27, 2007 Calculate the maximum speeds in both directions. (You should be able to determine that this is the value you get when the ball strikes the floor)
ender7x77 Posted February 27, 2007 Author Posted February 27, 2007 Thanks for the quick response. My initial thought was that since the ball never loses energy that it will start at 10m/s, then as it hits the time axis will become 0m/s, and then as it is descending it will be -10m/s, raising back again to zero to the impact, and then upwards to 10m/s. Is this a right way to look at it or am I completely wrong to make such speculations?
swansont Posted February 27, 2007 Posted February 27, 2007 Thanks for the quick response. My initial thought was that since the ball never loses energy that it will start at 10m/s, then as it hits the time axis will become 0m/s, and then as it is descending it will be -10m/s, raising back again to zero to the impact, and then upwards to 10m/s. Is this a right way to look at it or am I completely wrong to make such speculations? You are not accounting for the fact that it starts 2m off the ground, and will have more KE when it hits the ground than when you launch. The ground, where it has minimal PE, is where it has maximum KE, since the sum of the two is constant.
ender7x77 Posted February 27, 2007 Author Posted February 27, 2007 Ya, I figured that much just forgot to include it because class was about to begin. The problem is that i have to make calculations in order to make a scale, but I do not know how to approach doing it. Also, I took the class without any prerequisites so my understanding of things is a bit rough. thanks.
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