Solving Ball Dropped in Accelerating Elevator: 2.1m, 4.6m/s, 1.2m/s2

In summary: The ball and the elevator floor will be at the same height above the ground when the ball hits the floor.
  • #1
Naeem
194
0
Ball in Elevator...

Q.
Ball in Elevator

--------------------------------------------------------------------------------

A physicist is inside an elevator rising in a skyscraper. She is holding a ball at a height h = 2.1 m above the floor of the elevator.

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a) She let's go of the ball while the elevator is rising at a constant speed of 4.6 m/s. How long does it take the ball to hit the elevator floor?
tto floor = s
2.1/4.6 NO

HELP: The ball and the elevator floor will be at the same height above the ground when the ball hits the floor

I used

x - x0 =v0t + 1/2at^2

Since constant speed acc. is zero.

so,

t = x-xo/vo
2.1 /4.6 which is wrong?

Any one help
--------------------------------------------------------------------------------
b) Now let's try a somewhat harder variation on this problem. Suppose instead that the elevator is accelerating upward with constant acceleration a = 1.2 m/s2 and that the upward speed of the elevator at the instant the ball is dropped, is 4.6 m/s. Now how long does it take the ball to hit the elevator floor, assuming once again that it is dropped from a height of 2.1 m? Q


No idea here help me, what equations, may be the same one as above, what to plugin, etc.etc.
 
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  • #2
The acceleration of the elevator is zero, the acceleration due to gravity is still 9.8 m/s. For both parts you need to write separate position equations for both the elevator and ball, and solve for when they are equal.
 
  • #3
Naeem said:
Q.
Ball in Elevator

--------------------------------------------------------------------------------

A physicist is inside an elevator rising in a skyscraper. She is holding a ball at a height h = 2.1 m above the floor of the elevator.

--------------------------------------------------------------------------------
a) She let's go of the ball while the elevator is rising at a constant speed of 4.6 m/s. How long does it take the ball to hit the elevator floor?
tto floor = s
2.1/4.6 NO

HELP: The ball and the elevator floor will be at the same height above the ground when the ball hits the floor

I used

x - x0 =v0t + 1/2at^2

Since constant speed acc. is zero.

so,

t = x-xo/vo
2.1 /4.6 which is wrong?

Any one help

The ball is falling to the floor of the elevator and the elevator is rising to meet the ball. Think of two particles moving towards each other. Write their positions over time and solve them simultaneously to find when they intersect.
For the ball

[itex]y = y_0 -\frac{1}{2}gt^2[/itex]

like you said. For the floor of the elevator

[itex]y_{\mathrm{fl}} = y_{\mathrm{fl},0}+ v_{\mathrm{fl}} t.[/itex]

Naeem said:
b) Now let's try a somewhat harder variation on this problem. Suppose instead that the elevator is accelerating upward with constant acceleration a = 1.2 m/s2 and that the upward speed of the elevator at the instant the ball is dropped, is 4.6 m/s. Now how long does it take the ball to hit the elevator floor, assuming once again that it is dropped from a height of 2.1 m? Q


No idea here help me, what equations, may be the same one as above, what to plugin, etc.etc.

Same thing as above, except you need to include the [itex]1/2 at^2[/itex] term in the second equation to take into account the elevator's upward acceleration.
 
  • #4
Yep! Got em all! Thanks,
 
  • #5
hello! I have the same problem only with different values:

the height of the ball is h = 1.2 m instead.

a) She let's go of the ball while the elevator is rising at a constant speed of 4.7 m/s. How long does it take the ball to hit the elevator floor?

what i did was as listed above i did:

[itex] y = 1.2 - \frac{1}{2}(9.8)t^2[/itex] for the ball and

[itex]y_{\mathrm{fl}} = 0 + 4.7t[/itex] for the elevator

i graphed those two and the intersection is at t = .2095

why is this not the right answer? thanks!
 
Last edited:

Related to Solving Ball Dropped in Accelerating Elevator: 2.1m, 4.6m/s, 1.2m/s2

What is the formula for calculating the final velocity of the ball when dropped in an accelerating elevator?

The formula for calculating the final velocity of the ball is v = u + at, where v is the final velocity, u is the initial velocity (in this case, 0 m/s), a is the acceleration (1.2 m/s2), and t is the time (in this case, the time taken for the ball to drop from 2.1 m, which can be calculated using the formula t = √(2h/a), where h is the initial height of the ball).

What is the final velocity of the ball after being dropped in an accelerating elevator?

The final velocity of the ball can be calculated using the formula v = u + at, where u is the initial velocity (in this case, 0 m/s), a is the acceleration (1.2 m/s2), and t is the time taken for the ball to drop from 2.1 m, which can be calculated using the formula t = √(2h/a), where h is the initial height of the ball. Plugging in the values, we get v = 0 + 1.2(√(2(2.1)/1.2)) = 4.6 m/s.

How does the acceleration of the elevator affect the final velocity of the ball?

The acceleration of the elevator affects the final velocity of the ball because it determines how quickly the ball will gain velocity as it falls. In this case, the acceleration of the elevator is 1.2 m/s2, so the ball will gain 1.2 m/s of velocity every second it falls. This results in a final velocity of 4.6 m/s after the ball has dropped 2.1 m.

What would happen if the acceleration of the elevator was 0?

If the acceleration of the elevator was 0, this would mean that the elevator is not moving and there is no change in velocity. In this case, the ball would simply fall at a constant speed of 9.8 m/s2, as determined by the force of gravity. Therefore, the final velocity of the ball would be 9.8 m/s after it has dropped 2.1 m.

Can this same calculation be applied to objects of different masses?

Yes, this calculation can be applied to objects of different masses as long as the only force acting on the object is gravity and the object is dropped from rest. The acceleration due to gravity (9.8 m/s2) would remain the same, but the final velocity would vary depending on the mass of the object. Heavier objects would have a greater final velocity than lighter objects after being dropped from the same height in an accelerating elevator.

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