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Sciene fair project, freefall stuff - need one equation for time |
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| Jan22-12, 11:32 AM | #1 |
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Sciene fair project, freefall stuff - need one equation for time
1. The problem statement, all variables and given/known data
Ok, so I'm doing this science fair project on tennis ball bounce hieght, and I have to find the force that it hits the ground with. I'm trying to do this mathematically, because I can't record the time accurately in the actual experiment. The mass of the ball is 0.0567kg, the acceleration is just gravity, 9.8m/s/s, and the height of the drop varies; 0.2m, 0.4m, 0.6m, and 1m. I just need to know how to calculate time, but I'm not sure what the equation is. I tried t=√2d/a, but that gives me very wierd results. 2. Relevant equations t=√2d/a (?) v=d/t F=m(Δv/Δt) 3. The attempt at a solution so for the 0.2m height, I did these calculations. d=0.2m a=9.8m/s/s m=0.0567kg t=√2d/a t=√0.4m/9.8m/s/s t=√0.40s/s t=0.2s v=d/t v=0.2m/0.2s v=1m/s F=m(Δv/Δt) F=0.0567kg(1m/s /0.2s) F=0.0567kg(5.0m/s/s) F=0.284N and for the 0.4m height, I did d=0.4m a=9.8m/s/s m=0.0567kg t=√2d/a t=√0.8m/9.8m/s/s t=√0.08s/s t=0.894s v=d/t v=0.4m/0.894s v=0.447m/s but that doesnt make sense! how can the velocity be less from a greater height? I feel like I'm doing the time equation wrong, and I've looked everywhere online and can't find a reliable time from distance and acceleration equation. This is driving me crazy, please help! Also, am I calculating impact force correctly? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution |
| Jan22-12, 11:40 AM | #2 |
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Welcome to PF!
The first part where you find t = 0.2 s is correct. After you use v = d/t things go wild because v= d/t does not apply to accelerated motion. Use v = a*t to find the final velocity. Air resistance may be significant with a tennis ball. Have you thought of taking a movie of the bouncing ball with a clock or stopwatch in the picture? You should be able to get a pretty accurate record of the distance and time that way. |
| Jan22-12, 11:45 AM | #3 |
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Don't confuse the time it takes for the ball to drop (which is easy to calculate using freefall equations) with the duration of the collision with the ground (not so easy to calculate). You'll have to poke around to see if you can get estimates of the latter.
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| Jan22-12, 11:54 AM | #4 |
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Sciene fair project, freefall stuff - need one equation for timeAre you sure that the air resistance would be significant at this small of a height? |
| Jan22-12, 12:47 PM | #5 |
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Hint:
You have already computed the time it takes for the ball to reach the ground. If you were to measure how high the ball bounces, you can determine the time it took to reach that altitude. Now if you timed the entire trip, you can perform some other calculations that would lead you to an estimate of the average force on the ground. |
| Jan22-12, 12:59 PM | #6 |
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| Jan22-12, 01:08 PM | #7 |
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To find the force during the collision, you need the time of the collision. LawrenceC has an interesting idea for measuring that collision time, but I suspect it won't work in practice since the collision time will be too small. But worth a shot! |
| Jan22-12, 02:02 PM | #8 |
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| Jan22-12, 02:55 PM | #9 |
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I wonder, could I use F=(mv)/t of this?
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| Jan22-12, 03:40 PM | #11 |
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| Jan22-12, 04:07 PM | #12 |
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I'm going to calculate KE, and then force from that
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| Jan22-12, 04:10 PM | #13 |
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| Jan22-12, 04:24 PM | #14 |
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Posting new calculations for 0.2m
d=0.2m a=9.8m/s/s m=0.0567kg t=√2d/a t=√0.4m/9.8m/s/s t=√0.04s/s t=0.2s v=at v=9.8m/s/s*0.2s v=1.96m/s KE=1/2 m v^2 KE=0.028kg*3.841m^2/s^2 KE=0.108J |
| Jan22-12, 04:30 PM | #15 |
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Could I also base force calculations off of the distance travelled into the ground after collision? I'd estimate that to be around 0.0001m on a hardwood floor, just a guess. |
| Jan23-12, 11:29 AM | #16 |
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You should be able to SEE the compression of the ball on a video of it hitting the floor. You could repeat it a few times to get a better picture. Even a cell phone video might be good enough. Then measure what force is needed to compress the ball to the same compression you see in the video. Measurements are needed because theoretical methods depend on the nature of the ball.
I think a measurement of the duration of the ball's collision with the floor would be more useful. From that, you could use the impulse formula to calculate the average force. |
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| acceleration, distance, science fair, time, urgent |
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