Help with free falling object problem

  • #1
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.
 

Answers and Replies

  • #2
1,860
0
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
318
0
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
66
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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
74
0
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
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.
 

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