Relative Velocity

A lift is moving up with an acceleration of 2 m/s 2 2\text{ m/s}^2 . A main in it throws a ball up with a velocity of 12 m/s 12\text{ m/s} with respect to himself. The ball comes back to the man in time t t . If the ball doesn't hit the roof of the lift, then find t t .

Take g = 10 m/s 2 g = 10\text{ m/s}^2 .

None of these 3 seconds 1 second 2 seconds

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Swagat Panda
May 27, 2015

Let the lift’s velocity be v. Now it is obvious that \text{Let the lift's velocity be v. Now it is obvious that}\rightarrow

Displacement of the ball= Displacement of the man \text{Displacement of the ball= Displacement of the man}

( v + 12 ) t 1 2 g t 2 = v t + 1 2 a t 2 where a is the acceleration of the lift. \Rightarrow (v+12)t-\frac { 1 }{ 2 } gt^{ 2 }=vt+\frac { 1 }{ 2 } at^{ 2 } \text{where a is the acceleration of the lift.}

( v + 12 v ) t = 1 2 t 2 ( a + g ) 12 t = 6 t 2 t = 2 seconds \Rightarrow (v+12-v)t=\frac { 1 }{ 2 } t^{ 2 }(a+g)\\ \Rightarrow 12t=6t^{ 2 }\\ \Rightarrow t=2 \text{ seconds}

it's given in vmc supplementary , right?

A Former Brilliant Member - 4 years, 10 months ago

I did this from somewhere else, maybe it is also in the supplementary, I am not too sure.

Swagat Panda - 4 years, 10 months ago

Log in to reply

don't you attend class regularly?

A Former Brilliant Member - 4 years, 10 months ago

I think sir discussed it in the class or something like that, I don't remember exactly, and I am not going to the class from the last few days because I was ill and I have to see the videos of previous classes. Did you not go to the class today?

Swagat Panda - 4 years, 10 months ago

Log in to reply

no i went to the class , but i saw that you commented during the class - time , due to which i deduced that you missed the class. also which discussion are you talking about ?

A Former Brilliant Member - 4 years, 10 months ago

Oh, Okay. Are you on any social network because I feel a little awkward talking here in comments?

Swagat Panda - 4 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...