
An elevator moves upward with an upward acceleration of \[1m{s^{ - 2}}\]. At an instant when the velocity is $2m{s^{ - 1}}$, a ball is thrown with initial velocity of $2m{s^{ - 1}}$ with respect to the elevator at an elevation of ${30^ \circ }$. If the elevator is moving with a downward acceleration of g how will the motion be altered?
Option A: straight line w.r.t to elevator and projectile w.r.t ground.
Option B: projectile with respect to elevator and straight w.r.t ground.
Option C: projectile w.r.t both.
Option D: straight w.r.t both.
Answer
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Hint:The concept of relative motion is going to be used in this problem. The velocity of a body with respect to another moving body differs from its true velocity by a factor of the other body’s velocity.
Complete solution:
Here the case is very simple: the elevator is moving downwards with a velocity of $2m{s^{ - 1}}$ and it has a downward acceleration of g so the elevator is undergoing a free fall motion.
Now when the ball is thrown from the elevator; the velocity of the elevator is $2m{s^{ - 1}}$ downwards.
Now let us talk about the ball,
The ball is thrown in the upward direction at an angle of ${30^ \circ }$ from the elevator surface.
Also the velocity at that point of the ball is $2m{s^{ - 1}}$ with respect to the elevator.
But the acceleration of the ball is ‘g’ downwards so the relative acceleration of both the bodies is zero.
Thus both bodies undergo a free fall motion with zero relative acceleration.
Thus the trajectory of the ball will be a straight line with respect to the elevator.
However, for someone on the ground frame i.e. outside the elevator the ball will look like it is undergoing a projectile motion.
Therefore option A is correct.
Note:
-If the relative acceleration was not zero then the motion would have been a projectile in both the cases.
-For a better understanding you must try and draw a diagram for the given case.
-In this case both the bodies appear to have been undergoing free fall with zero acceleration with respect to each other, so the relative motion was like a straight line.
Complete solution:
Here the case is very simple: the elevator is moving downwards with a velocity of $2m{s^{ - 1}}$ and it has a downward acceleration of g so the elevator is undergoing a free fall motion.
Now when the ball is thrown from the elevator; the velocity of the elevator is $2m{s^{ - 1}}$ downwards.
Now let us talk about the ball,
The ball is thrown in the upward direction at an angle of ${30^ \circ }$ from the elevator surface.
Also the velocity at that point of the ball is $2m{s^{ - 1}}$ with respect to the elevator.
But the acceleration of the ball is ‘g’ downwards so the relative acceleration of both the bodies is zero.
Thus both bodies undergo a free fall motion with zero relative acceleration.
Thus the trajectory of the ball will be a straight line with respect to the elevator.
However, for someone on the ground frame i.e. outside the elevator the ball will look like it is undergoing a projectile motion.
Therefore option A is correct.
Note:
-If the relative acceleration was not zero then the motion would have been a projectile in both the cases.
-For a better understanding you must try and draw a diagram for the given case.
-In this case both the bodies appear to have been undergoing free fall with zero acceleration with respect to each other, so the relative motion was like a straight line.
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