Solving the Falling Ball Problem: t2=2t1?

In summary, two students standing on a fire escape, one twice as high as the other, simultaneously drop a ball. The question is when will the second ball strike the ground, given that the first ball strikes at time t1. The options for the time t2 are A) t2=4t1, B) t2=5/4t1, C) t2=2t1, and D) t2=squareroot of 2 times t1. To solve this problem, we can use the equation d=-1/2*g*t^2, where d is the distance, g is the acceleration due to gravity, and t is time. By plugging in the values for t2 and
  • #1

Homework Statement


two students are standing on a fire escape, one twice as high as the other. simultaneously, each drops a ball. if the first ball strikes the ground at time t1 when willt he second ball strike the ground?
A)t2=4t1
B)t2=5/4t1
C)t2=2t1
D)t2=squareroot of 2 times t1


Homework Equations





The Attempt at a Solution


i tried plugging in numbers and well its two times the height so wouldn't it be C??
 
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  • #2
Yay I get to use my skydiving experience! When jumping out of an airplane I know it'll take roughly 10 seconds to fall the first 1000 feet, then roughly 5 seconds for every subsequent 1000 feet. So 10 seconds for 1000 feet, 15 seconds for 2000 feet(pretty gross estimation, but it overestimates how far you've fallen usually which is better than underestimating for obvious reasons!)

So think about the equation you should be using here

d=-1/2*g*t^2

What you've said in thinking it's C is that if I have a ball that falls a distance D1 in T1 seconds then if T2=2*T1, then D2=2*D1

is that true?
 
  • #3
wen i tried pluging in numbers i kept the initial velocity, aceleration the same and made one distance x and the other one 2x and then i tried finding the final velocity. then using one of the formulas i found how much time it would take and compared them. so the Vi and A is the same but the D and Vf are different.
 
  • #4
You'll confuse yourself trying to plug in numbers

so D=-1/2*g*T^2, 1/2 and g are constants

Now we're saying it falls twice D in time T2, so 2*D=-1/2*g*T2^2, and you have said that T2=2*T, or that's what choice C says anyways. So what you're saying is 2*D=-1/2*g*(2*T)^2

You KNOW that D=-1/2*g*T^2 from our very first step, so if you solve the bolded equation for D, do you get the same thing?
 
  • #5
yea the both equation do equal out. but how would that prove that the time would be 2 times if the distance is 2x?
 

1. How do I solve the falling ball problem using t2=2t1?

The first step in solving the falling ball problem using t2=2t1 is to understand the equations involved. The equation t2=2t1 represents the relationship between the time taken for an object to reach its maximum height (t1) and the time taken for the object to return to its original height (t2). By manipulating this equation, you can solve for other variables such as initial velocity and acceleration.

2. What are the assumptions made when using t2=2t1 to solve the falling ball problem?

When using t2=2t1 to solve the falling ball problem, it is assumed that the object is in free fall and that there is no air resistance or other external forces acting on the object. It is also assumed that the acceleration due to gravity is constant throughout the fall and that the initial velocity is zero.

3. Can t2=2t1 be used for any falling object?

No, t2=2t1 can only be used for objects in free fall. This means that the object must be only influenced by the force of gravity and not any other external forces. If there is air resistance or other external forces present, t2=2t1 may not accurately represent the falling object's motion.

4. How can I use t2=2t1 to find the maximum height reached by a falling ball?

To use t2=2t1 to find the maximum height reached by a falling ball, you can rearrange the equation to solve for the initial velocity (v0). Once you have the initial velocity, you can use the equation h = v0*t - 1/2*g*t^2, where h is the maximum height, v0 is the initial velocity, t is the time taken to reach the maximum height, and g is the acceleration due to gravity. Plug in the values you have and solve for h.

5. Are there any limitations to using t2=2t1 to solve the falling ball problem?

Yes, there are limitations to using t2=2t1 to solve the falling ball problem. As mentioned before, it can only be used for objects in free fall and may not accurately represent the motion of objects with air resistance or other external forces. Additionally, it assumes a constant acceleration due to gravity, which may not be the case in real-life scenarios where the object may encounter different gravitational fields.

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