Question

An elevator is going up with an upward acceleration of 0.2m/s^2 . At the istant when the velocity is 0.2m/s a stone is projected upwards from its floor with a speed of 2m/s at an angle 30 degree with horizontal, relative to the elevator.
Find (a) The time taken by the stone to return to the floor

(b) The distance travelled by the floor of the lift from the time the stone was projected to the time when the stone hits the floor is?

Answer 1

The answer is (a) 0.2s (b) 0.404m

Answer 2

relative initial v = 0.2 m/sec
relative acc. = -g-0.2 m/sec^2


cal. final vel. now using newton's equations of motions.

Answer 3

Since it was thrown with an angle why we need to consider the acceleration

Answer 4

T=2usintheta/g can we use that??

Answer 5

hey is time taken =1sec

Answer 6

sorry i mean 2sec ??

Answer 7

and yea shakir u can use dat formula u mention above

Answer 8

How in question it is given that theta=30

Answer 9

lol am sorry I didn't read the question carefully. @heena will help you out :D

Answer 10

@heena i got the Time of Flight Plzz help me to do distance

Answer 11

so u knw already lift imoving means tone already have velocity 2m/s with it now u also thrown it by 2m/s velocity so total velocity becomes 4m/s
now we know qn is like this

here u knw acceleration of lift is 0.2m/s^2 means it will be same for stone too now u have u=4m/sec u have theta 3odegree and here g will be taken as 0.2m/s^2 nw put the values in formula u ll get it