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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 ?


4 years ago

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 = gassume 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/ghence 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 = uConstant velocity of the balloon = -vLet 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 $\dpi{80} {\color{Red} s = ut+\frac{1}{2}gt^{2}}$ , we get                     $\dpi{80} {\color{Red} -vt = ut-\frac{1}{2}gt^{2}}$     Solving, we get $\dpi{80} {\color{Blue} t= \frac{2(u+v)}{g}}$ , 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 + vNow 0=-(u+v)+gtwhich gives, t=u+v/gbut this isn’t the time of flight.We know that time of flight, T=2t=2(u+v)/gPerhaps that’s the right answerhere’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 wrongMeant to say we cannot solve using-vt=ut-1/2gt2Though the answer would be the same but this isn’t the correct way, not at all

3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions