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 ?

Question

RANJEET KUMAR · Answer

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 !

Kuldeep Pal · Answer

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.

RANJEET KUMAR · Answer

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.

Raman Mishra · Answer

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.

RANJEET KUMAR · Answer

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.

Raman Mishra · Answer

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

Niraj Bishnoi · Answer

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 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

Niraj Bishnoi · Answer

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

Niraj Bishnoi · Answer

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

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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 ?

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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