Short Webpage Title: Ball Toss Free Fall Equations

AI Thread Summary
The discussion focuses on a physics problem involving two students throwing balls from a balcony, requiring symbolic solutions for time intervals, velocities, and distances. The first part of the problem involves calculating the time difference between when each ball strikes the ground, with initial equations provided for each ball's motion. Participants clarify that the time variables for each ball must be treated separately and emphasize the need for correct application of kinematic equations. For the second and third parts, suggestions are made to use different kinematic equations to find the velocities and distances at a given time. The conversation highlights the importance of careful equation setup and solving quadratic equations for accurate results.
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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