A glass marble dropped from a certain height above the horizontal surface reaches the surface in time t and then continues to bounce up and down. The time in which the marble finally comes to rest is (where e is the coefficient of restitution)

A glass marble dropped from a certain height above the horizontal surf

1.	A glass marble dropped from a certain height above the horizontal surface reaches the surface in time t and then continues to bounce up and down. The time in which the marble finally comes to rest is (where e is the coefficient of restitution)
Answer» `E^(n)t` `e^(2)t` `t[(1 + e)/(1 - e)]` `t[(1 - e)/(1 + e)]` Solution :Time TAKEN by glass marble to fall through height h is `t = sqrt((2h)/(g))`. After first REBOUND, time taken to rise to height `h_1` and COMING BACK `= 2sqrt((2h_1)/(g)) = 2sqrt((2e^2h)/(g)) = 2e sqrt((2h)/(g))` Time taken in rising to height, `h_2` and coming back = `2e^2 sqrt((2h)/(g))` Total time taken by the glass marble in coming to rest `= sqrt((2h)/(g)) + 2e sqrt((2h)/(g)) + 2e^2 sqrt((2h)/(g)) + .........` `= sqrt((2h)/(g)) + 2e sqrt((2h)/(g)) [1 + e + e^2 + ........]` `= sqrt((2h)/(g)) + 2e sqrt((2h)/(g)) xx 1/(1 - e) = sqrt((2h)/(g)) [1 + (2e)/(1 - e)] = t [(1 + e)/(1 - e)]` .