Recent content by Salk13

it is 7.93 m/s since the marble is going up and the elevator is going down

i know 10.03 m/s is relative to Earth but i don't know how it is different after the bounce

Bvu it should be from the ground which is the glass plate haruspex -10.03 m/s

height is relative to the point where the marble first dropped but how do you find the height (the answer) is there another way cause I am not seeing how to get this answer?

yes I was talking about the height as in answer to question

10.03 m/s because the collision is elastic and kinetic energy is conserved based on that would height be 5.127m ?

The relative velocity would be 10.03 m/s

it would be 0. I would use the equation Vac= Vab+Vbc with a= marble, b=elevator and c= ground. If they are both falling to the ground, then Vac=Vbc=2.1 m/s, leaving the relative velocity to 0.

14.23 m/s

they are traveling in the same direction

I got 12.13 m/s relative to the shaft. Then, using relative motion, the marble should be 14.23 m/s relative to ground.

BvU sorry for the confusion on equation I meant to say v[2] = u[2] + 2(a)x with x=7.5 u= speed of elevator and v= velocity of ball before impact haruspex okay I thought about finding the speed of the marble before impact by using conservation of energy. I am having trouble with getting the...

Homework Statement [/B] A glass marble is dropped down an elevator shaft and hits a thick glass plate on top of an elevator that is descending at a speed of 2.10 m/s. The marble hits the glass plate 7.5 m below the point from which it was dropped. If the collision is elastic, how high will the...