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Free fall, linear regression please help

  1. Sep 11, 2007 #1
    1) To test the quality of a tennis ball you drop it onto the floor from a hieght of 4 m. it rebounds to a hieght of 2 m. if the ball is in contact with the floor for 12 ms, what is the magnitude of its average acceleration during contact and is the average acceleration up or down.

    What i did was use the formula x=xi+vit+.5at^2. xi=4, x=2, vi=0, t=.012 s. I got the answer to be like 27,000. I know that is wrong because the answer is suppose to be 1,260. could you help me get in the right direction like what i need to find first if i need the acceleration or if i maybe used wrong infoormation. Also for the second question is that shown by positive or negitive acceleration.

    2) A parachutist bails out and freely falls 50m. then the parachute opens, and thereafter she decelerates at 20m/s^2. She reaches the ground with a speed of 3.0 m/s. How long is the parachutist in the air? at what height does the fall begin.

    For this one i attempted to use the formula V^2-vi^2+2ax. I got some rediculous small hieght. I used a=-20, v=3, vi=0. Once agian could you tell me if there is a varible i need to find first to get me satrted in the right direction.

    3) This isnt the whole problem but this is the information needed to solve my question about linear regression. here are the points on a graph as such d-t. 0-0, 10-1.63, 20-2.33, 30-2.83, 40-3.31, 50-3.79. The question is use linear regression of the graph to find the magnitude of acceleration. You had to graph the points d versus t^2.

    When graphing my points i just did t^2 and got 2^2 to equal 2.67-5.73-8-11-14.36. I dont rally know what linear regression is. i tried to solve it logically. i used the formula x-vit+.5at^2. Since it is at rest intially velocity is 0. I ended up with x=.5at^2. I solved for a and got a=(2x)/(t^2). I plugged in the numbers and got accelerations of 7.53, 7.37,7.5,7.27,6.96. I averaged them together and get 7.4. the answer is suppose to be 7.2 is this just cause i did not use sig figs or did i solve it the wrong way.

    Thank you so much for all your help i really find this forum to be very helpful in physics
  2. jcsd
  3. Sep 11, 2007 #2

    D H

    Staff: Mentor

    The ball falls just shy of 4 meters before it hits the floor. It has some downward velocity at the instant it first touches the floor. 0.12 seconds later, it has enough upward velocity to make it rise 2 meters above the floor. Calculate these two velocity vectors. The difference between these two vectors (keep in mind that one is downward and the other upward) is what you need to work with.

    Problem 2. Break this problem into two parts: the part freefall (acceleration is downward) and the part with the parachute open (acceleration is upward) . You assumed a constant acceleration, which is not valid.
  4. Sep 11, 2007 #3
    Dh, for the first problem i got velcoty of vector 1 to be 8.85 and velocity of vector 2 to be 6.26. I subtracted them and got 2.59. I used the formula v=vi+at and i still got the wrong answer i used v=2.59, vi=0, a=? t=.012.
    For the second problem i split it into time of fall first 50m and time of fall using parachute. for the free fall part i got a time of 3.19 and for the second time i got 1s. that cannot be right. first i found the final velocity at the 5o m free fall to be 22 m/s and than i used the formula v=vi+at using v=3 , vi=22, a=20. what am i doing wrong
  5. Sep 11, 2007 #4

    D H

    Staff: Mentor

    For the first problem, you are doing the vector difference incorrectly. Your vector 1 is directed downward, vector 2 upward. You are subtracting their magnitudes, which is incorrect.

    For the free-fall, you got the right time (3.19 sec) but not the velocity (what is 9.8*3.19?). The correct velocity will change your answer a bit. The answer is admittedly silly, but this is because in this problem the parachutist decelerates all the way from chute-open to the ground. Are you sure you have the problem right? Parachutists typically reach a small terminal velocity shortly after the chute opens. In other words, the acceleration profile is 9.8 m/s^2 downward during free-fall, a large acceleration upward for a short time, and then a constant velocity until touchdown.
  6. Sep 11, 2007 #5
    thanks D H i was able to do both problems correctly Thanks for all your help.
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