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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 ?

Grade:12th pass

9 Answers

RANJEET KUMAR
32 Points
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 +1/2gt2 =vt
=> t[gt/2 – u] =0
=> t = 2u/g
hence required time is 2u/g.
thanku ! 
Kuldeep Pal
53 Points
6 years ago
Dear Ranjeet Kumar,
How can you Take Displacement of ball be “vt” .
It should be “vt+1/2gt2.
Its right Answer is 2(u+v)/g.
I know the answer. But dont know how to solve.
RANJEET KUMAR
32 Points
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.
Raman Mishra
67 Points
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 {\color{Red} s = ut+\frac{1}{2}gt^{2}} , we get
                     {\color{Red} -vt = ut-\frac{1}{2}gt^{2}}
     Solving, we get {\color{Blue} t= \frac{2(u+v)}{g}} , which is the required answer.
RANJEET KUMAR
32 Points
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 t = 2[u+v]/g{\color{Blue} }, but if we are taking initial velocity to be ‘u’ with respect to balloon then the answer will be t = 2u/g. 
Raman Mishra
67 Points
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.
Niraj Bishnoi
50 Points
5 years ago
I’m in a shock that a 12th pass student has asked me this question when a 11th 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)gt2 [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
50 Points
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
Niraj Bishnoi
50 Points
5 years ago
And equaling the displacement of the ball and balloon would be wrong
Meant to say we cannot solve using
-vt=ut-1/2gt2
Though the answer would be the same but this isn’t the correct way, not at all

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