Help with free falling object problem

In summary: Can you help me out?In summary, the question is asking how many seconds apart two balls should be released in order for a ball dropped from a height of 5H to reach the ground at the same time as a ball dropped from a height of H. The answer is that both balls would need lift and the equation of free fall would need to be applied to determine the ratio of the lift.
  • #1
mathphysicnb
6
0
I hope this is the right forum, so I apologize in advance if I'm mistaken.

Here's the question poised:

How many seconds apart should two balls be released so that a ball being dropped from a height of H reaches the ground at the same time as a ball dropped from a height of 5H?

I need some help on this one. Thank you.
 
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  • #2
Uhhh, what kind of drag are we supposed to assume? Are you supposed to give one an initial velocity?

I don't think this is possible even with drag, the ball from H would need lift too. Unless H is something like 1000m and the drag is nonlinear from a really massive object, would that even do it...?
 
  • #3
Welcome to the forums.

First thing you must show your attempt. Whatever it may be.

Now for the question, are you familiar with the equations of kinematics.
 
  • #4
Its certainly physically possible, but he's probably missing some info.
 
  • #5
nope, that's the question. I'm fairly sure it involves finding a ratio, due to the many unknowns, so I imagine finding the value of H is not the goal. It's also assumed to be no air resistance. The two balls are released or dropped, so Vo = 0 m/s for both balls.

Show some work? OK.

Vi1 = Vi2 = 0 m/s

Xi1 = H; Xi2 = 5H; Xf1 = Xf2 = 0 m

ti1 = ti2 = 0 s

a = -9.8 m/s^2 = g

tf1 = ? ; tf2 = ? ; Vf1 = ? ; Vf2 = ? ; tf2-tf1 = ?

...

If I knew H, I could solve for tf1, tf2, and find the difference of the two

O --> ball with Xi2 = 5H
| |
| |
| | --> With H, I can find this difference: tf2 - tf1
| |
| |
| O --> ball with Xi1 = H
|
|
|
---------------------- ground, 0 meters


Xf1 = Xi1 + Vi1*ti1 + 0.5g*tf1^2

0 = H + 0 + 0.5g*tf1^2 ==> H = -0.5g*tf1^2 ==> tf1 = sqrt(-2H/g)

Xf2 = Xi2 + Vi2*ti2 + 0.5g*tf2^2

0 = 5H + 0 + 0.5g*tf1^2 ==> H = -10g*tf2^2 ==> tf2 = sqrt(-10H/g)

tf2-tf1 = ?

Finding the ratio of tf2/tf1 makes no sense and I know it's 5. If i square tf2-tf1, I'm changing the expression. So, at this point, I'm a bit stuck...
 
  • #6
does anybody have any suggestions or anything at all?
 
  • #7
now here is the equation of free fall

[tex] X(t) = X0 - V0 * t - \frac{g*t^2}{2}[/tex]

apply it to the two ball, remember that you have a delay between the two droppings.

-----------------------------------------------------
Correct me if I am wrong.
http://ghazi.bousselmi.googlepages.com/présentation2[/QUOTE]
 
  • #8
tabchouri said:
now here is the equation of free fall

[tex] X(t) = X0 - V0 * t - \frac{g*t^2}{2}[/tex]

apply it to the two ball, remember that you have a delay between the two droppings.

-----------------------------------------------------
Correct me if I am wrong.
http://ghazi.bousselmi.googlepages.com/présentation2
[/QUOTE]

Um, did you read my post at all? I know this formula and I know the difference between the two times is what I'm looking for. The problem is I'm stuck because the fact that I don't have a specific value for the height complicates things.

I know that tf2 = sqrt(5)*tf1. But I can't figure out a way to use this to my advantage.
 

1. What is a free falling object?

A free falling object is an object that is only being affected by the force of gravity. This means that there are no other forces acting on the object, such as air resistance or friction.

2. How can I calculate the distance traveled by a free falling object?

To calculate the distance traveled by a free falling object, you can use the formula d = 1/2 * g * t^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time the object has been falling.

3. What is the difference between free fall and constant acceleration?

Free fall is a type of motion where an object is only influenced by the force of gravity, while constant acceleration refers to any motion where the velocity of an object changes at a constant rate. Free fall is a type of constant acceleration because the acceleration due to gravity remains constant.

4. How does the mass of a free falling object affect its motion?

The mass of a free falling object does not affect its motion. According to Newton's Second Law of Motion, the force of gravity acting on an object is equal to its mass multiplied by the acceleration due to gravity. Therefore, an object with a larger mass will experience the same acceleration due to gravity as an object with a smaller mass.

5. Can air resistance affect the motion of a free falling object?

Yes, air resistance can affect the motion of a free falling object. As the object falls, it will encounter air resistance, which is a force that opposes the motion of the object. This can cause the object to reach a terminal velocity, where the force of air resistance is equal to the force of gravity, and the object will stop accelerating.

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