11You forgot that when the string breaks the ball has a upward velocity which has to be considered.
Consider a elevator accelerating upward with 2.4m/s2.
Inside the elevator there is a ball of mass 1kg tied to the ceiling using a massless string.
The elevator travels 30m up and then the string breaks and the ball is let loose.
Calculate the time taken by it to reach the floor of the elevator which is 5m away from the hanging ball.
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13 Answers
621/2 g. t2 = 2.4
t2 = 4.8g
v= √4.8 t
downward, displacemnt will be 35 m for the ball..
-35= √4.8 tt - 1/2 g t2
now solve for t
there iwll be one -ve root. ignore that :)
11Bhaiyya the displacement is not 2.4 it is 30m of the elevator with the ball intact to the ceiling. 2.4 is the acceleration of the elevator and after 30 m the string holding the ball breaks and the ball falls . The time according to me is 0.67seconds
Please tell me if i am wrong
3lets luk at i frm outside point of view ..........
√2(2.4)30 = v =initial vel. upwards ..........
1/2gt2 - vt = 5 - distance covered by base of elevater in dat time.......
1111I think that you have not considered the acceleration of the elevator and its velocity.
3""""""""""""lets luk at i frm outside point of view """""""""
is a key statement ... so i havnt included psudo force bhai.....
u can do it other way also...
3""""""distance covered by base of elevater in dat time"""""""""""
covers all that ....
11ok i got it I did not read "distance covered by the base of the elevator"
You r rite.
