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jperk980
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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 isn't 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 don't 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
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 isn't 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 don't 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