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Grade: 12th pass

                        

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 ?

4 years ago

Answers : (9)

RANJEET KUMAR
32 Points
							
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 ! 
4 years ago
Kuldeep Pal
53 Points
							
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.
4 years ago
RANJEET KUMAR
32 Points
							
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.
4 years ago
Raman Mishra
67 Points
							
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.
4 years ago
RANJEET KUMAR
32 Points
							
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. 
4 years ago
Raman Mishra
67 Points
							
Ohh...sorry for the typing mistake. It should read: ‘...the velocity u of the ball must be with respect to the ground’. Apologies.
4 years ago
Niraj Bishnoi
50 Points
							
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
3 years ago
Niraj Bishnoi
50 Points
							
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
3 years ago
Niraj Bishnoi
50 Points
							
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
3 years ago
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