Question

if a ball is thrown up while travelling in a lift( which is moving up with uniform velocity ) what is the time of return? here is it not wrong to take time of ascent = time of descent as the lift is moving up? can ne1 help

Answer 1

A lift moving with uniform velocity is an inertial frame of reference.

Answers to your questions will be the same as if the lift were at rest.

Answer 2

Yes, the times will be different and also the speed with which the ball hit the lift will be different(which is same in case of stationary lift), yet total time of flight remains same as that of stationary lift.

Answer 3

@vincent & Naveen thanks.but i am a bit confused...why inertial frame of reference applies to lift but not for a car driving with uniform velocity?

Answer 4

A car with uniform velocity is also an inertial frame.

Answer 5

then if a bullet is fired from car then its resultant velocity is sum of car's and bullet's velocity, but in case of ball thrown from lift ,lift velocity is not added..why?

Answer 6

Well, the answer to your question depends in which frame of reference you are computing displacement and velocities.

If you use Earth as frame, then you have to add the lift velocity to that of your ball.
You can solve the problem twice, using Earth or lift as frame, then the answer for the time travelled will be the same.

Answer 7

Vincent thanks,good insight..but when lift is going up then 'time of descent' will not be less compared to when lift is stationary (while throwing the ball up)? because the travel during descent is getting shortened by the upward movement of lift.

Answer 8

Once again these times will depend on which frame you consider.

Total time can be calculated without referring to time of ascent + time of descent.

Answer 9

If lift is considered as the frame then total time will be same for both stationery and moving lift or diffrent?

Answer 10

I do not understand your question.

Answer 11

Let \(u\) be velocity at which the ball is thrown wrt lift.
Let \(u_L\) be velocity of lift wrt ground.

In lift frame (quantities with index 1) :

time up and time down are: \(t_{1u}=t_{1d}=\Large \frac {u}{g}\)
and time of flight is: \(t_{1f}=t_{1u}+t_{1d}=\Large \frac {2u}{g}\)

In ground frame (quantities with index 2) :

time up and time down are: \(t_{2u}=\Large \frac {u+u_L}{g}\) and \(t_{2d}=\Large \frac {u-u_L}{g}\)

and time of flight is: \(t_{2f}=t_{2u}+t_{2d}=\Large \frac {2u}{g}=\normalsize t_{1f}\)

Answer 12

thanks vincent for the explanation.
my observation: time of ascent is more for ground frame compared to lift frame ...thank you

Answer 13

Q: a thief's car is fleeing at 30 m/s,
chasing car's velocity is 2m/s,a bullet fired from chasing car at 40m/sec
distance between cars 1000m

how long it will take the bullet to hit the thief's car?