Short Webpage Title: Ball Toss Free Fall Equations

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Homework Help Overview

The problem involves two students throwing balls from a balcony: one ball is thrown downward and the other upward, both at the same initial speed. The discussion focuses on determining the time interval between when the balls hit the ground, their velocities upon impact, and their separation at a given time before they strike the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations of motion for both balls and question the assumptions regarding their positions and timing. There is an exploration of how to correctly apply the kinematic equations to find the required time intervals and velocities.

Discussion Status

Some participants have provided hints and corrections regarding the setup of the equations, emphasizing the need to treat the timing for each ball separately. There is an ongoing exploration of different kinematic equations that may be more suitable for parts of the problem.

Contextual Notes

Participants note potential confusion regarding the timing of the balls and the need to clarify the equations used for each part of the problem. There is also mention of the need to ensure that the equations reflect the correct initial conditions and assumptions about the motion of the balls.

Ab17
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Homework Statement



Two students are on a balcony a distance h above the S street. One student throws a ball vertically downward at a speed vi ; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of vi, g, h, and t. (a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground? (b) Find the velocity of each ball as it strikes the ground. (c) How far apart are the balls at a time t after they are thrown and before they strike ground?

Homework Equations


Xf=xi + vt + 0.5at^2

3. Attempt solution

(a) Xf1 = h - vit - 0.5gt^2
Xf2 = vit - 0.5gt^2

0= h - Vit -0.5gt^2 (strike ground)
0 = Vit -0.5gt^2 (strike ground)

Therefore: Vit -05gt^2 = h - Vit - 0.5gt^2
t = h/2vi

Dont know what to do for (b) and (c) and not sure if the solution for (a) is even right
 
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Ab17 said:
Xf1 = h - vit - 0.5gt^2
OK.

Ab17 said:
Xf2 = vit - 0.5gt^2
Careful: They are both on the balcony.

Ab17 said:
Therefore: Vit -05gt^2 = h - Vit - 0.5gt^2
t = h/2vi
Careful: The times are different! The t on the left is not the same as the t on the right.

Hint: Solve each one separately for the time it takes to hit the ground.
 
Thank you I didnt realize both are on the balcony. So I should be using the time found in part a for part b and c
 
Ab17 said:
So I should be using the time found in part a for part b and c
Part a asks for the time difference, so you won't need that in the other parts. You'll need the same equations you used in part a to solve part c. For part b I would use a different kinematic equation altogether. (See if you can find one that meets your needs.)
 
Is this right?

Xf1 = h + Vi.t - 0.5gt^2
Xf2 = h - Vi.t -0.5gt^2

0 = h + Vi.t - 0.5gt^2
0 = h - Vi.t -0.5gt^2

t1 = -2h /2vi - gt
t2 = 2h / 2vi + gt

t2-t1 = 8hvi/ 4vi^2 - g^2t^2
 
Ab17 said:
t1 = -2h /2vi - gt
t2 = 2h / 2vi + gt
The t on the right is different in the two equations. The first is t1, the second t2.
I have no idea how you got the line after that.
Go back to the preceding pair of equations, the ones starting 0=h. Write them out properly, i.e. using t1 and t2 as appropriate.
There is quite a quick route from there, but if you can't spot it just solve those quadratic equations in the obvious way.
 

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