# A Ball is thrown verticall upward with a velocity ‘u’ from the balloon descending with a constant velocity ‘v’. The ball will pass by the balloon after time ?

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A Ball is thrown verticall upward with a velocity ‘u’ from the balloon descending with a constant velocity ‘v’. The ball will pass by the balloon after time ?

## 9 Answers

6 years ago

The question needs the basic understanding of motion to solve it.

considering the downward direction to be +ve and taking ground as the reference point, we have :

initial velocity of ball thrown upward using relative motion = –[u-v]

constant velocity of ballon = v

acceleration of thrown ball = g

assume the ball meets the balloon after time ‘t’

concept : the time when ball meets the balloon, both will have same displacement in the +ve direction according to our assumption of direction.

according to second law of kinematics, we have :

displacement of ball = displacement of balloon

-[u-v]t +1/2gt

^{2}=vt=> t[gt/2 – u] =0

=> t = 2u/g

hence required time is 2u/g.

thanku !

6 years ago

Dear Ranjeet Kumar,

How can you Take Displacement of ball be “vt” .

It should be “vt+1/2gt

^{2}.Its right Answer is 2(u+v)/g.

I know the answer. But dont know how to solve.

6 years ago

mr. kuldeep pal, it has been given in the question that the ballon is decending with a constant velocity ‘v’, so we can’t take any acceleration or decelaration.

6 years ago

Dear Kuldeep, the velocity u of the ball must be with respect to the balloon. Then, taking the upward direction as positive we have:

Initial velocity of the ball = u

Constant velocity of the balloon = -v

Let the ball meets the balloon after a time t.

Displacement of the balloon after time t = -vt , which is also the displacement of the ball.

Now, applying , we get

Solving, we get , which is the required answer.

6 years ago

Dear raman the gist of question is that if we take intial velocity of ball with respect to ground to be ‘u’, then the answer will be , but if we are taking initial velocity to be ‘u’ with respect to balloon then the answer will be t = 2u/g.

6 years ago

Ohh...sorry for the typing mistake. It should read: ‘...the velocity u of the ball must be with respect to the ground’. Apologies.

5 years ago

I’m in a shock that a 12

^{th}pass student has asked me this question when a 11^{th}moving student can answer this with 5 min. Sorry for acting oversmart. So here’s the solution-Relative velocity of all with respect to the baloon = u + v

Now 0=-(u+v)+gt

which gives, t=u+v/g

but this isn’t the time of flight.

We know that time of flight, T=2t=2(u+v)/g

Perhaps that’s the right answer

here’s another way to solve it:-

We equa

=> -vt = ut - (1/2)gt

=> -v = u - (1/2)gt

=> (1/2)gt = u + v

=> t = 2(u+v)/g

^{2}[Since in the case of balloon the acceleration is zero]=> -v = u - (1/2)gt

=> (1/2)gt = u + v

=> t = 2(u+v)/g

And please don’t ask me why there’s minus, okay fine I’ll answer- I use down as negative and upward as positive, so following the convention, v will be negative and u will be positive

5 years ago

I didn’t explain in the previous answer, so here’s the explanation:-

A projectile aquires the speed of the body, in the direction it’s going (the projectile’s direction). So the speed of the ball thrown would be u+v. (since it has it’s own speed as well)

So, Time of ascent=u+v/g …....(A)

Time of descent=u+v/g …...(B)

Time of flight= 2(u+v)/g (A+B)

No neeed to say thanks

5 years ago

And equaling the displacement of the ball and balloon would be wrong

Meant to say we cannot solve using

-vt=ut-1/2gt

^{2}Though the answer would be the same but this isn’t the correct way, not at all