Recent content by jbm1939

Thanks insightful, I'll fix that on my problem.

No he did not say anything about significant figures, I generally just round to the nearest whole number. The question just say how much energy would be absorbed in the collision, so that would be the total amount of kinetic energy correct? What is derived from 1/2mv2?

thank you 24forChromium! its getting to the end of the semester here and my brain is overwhelmed at the moment. this forum is awesome for an extra set of eyes to help you see what you've overlooked.

so for KE=.5mv^2 =.5(400/32.2)(29.3)^2 = 5332.24 so would this be the amount absorbed by the tree in the impact? their would be no KE after the collision because v=0, so the tree has to absorb all that energy? am i correct with this deduction?

Homework Statement A 400 lb quad hits a tree moving at 20 mph, if the time of impact is .02 seconds what would the average force be on the tree? Also how much energy would be absorbed in the collision? Homework Equations I was able to solve the first part of the question, the force on the...

Does anyone know if I found the first height correctly? Another student told me that you just take the orignal height times e to get that. That seems to simple to me but I could be wrong. So in this case it would be 0.6 x 5ft = 3ft

Homework Statement A ball weighing 0.2 lbs is dropped out the window of a pickup truck moving horizontally at 45 mph. If the C.O.R. (e) between the ball and the ground is equal to 0.6, find the height of the ball's first bounce and the horizontal distance it will travel before bouncing again...