Solving Parabolic Motion: Finding h & v0

[jesuslovesu]

Homework Statement

http://img524.imageshack.us/img524/6899/tempnb3.th.png

Homework Equations

y = y0 - 1/2gt^2 + voy*t

The Attempt at a Solution

Well there are two methods that I used for solving this, but I think only one is correct:

First, the correct method:
1) Find the time at the top of the parabola and use the distance from the hoop to determine the time.
12 = 5 - 16t^2 + v0*t*(5/13)
x = v0*(12/13)*t = 24
t = .433s

v0 is 60 ft/s

2) Find the time at y = 3 (to hit the second slope)
0 = 5-3 - 16t^2 + vo(5/13)*t
t = 1.524s

3) Find h
x = vo(12/13)*t = h + 24 --> h = 60.4 ft

I think everything I did is correct, however, my TA in lecture today said that the v0 should be between 50 ft/s - 60 ft/s and h should be between 40 ft and 47 ft. However, he also said that he didn't quite know how to get the correct solution... (he was looking at the solutions manual)

I get 60ft/s but my h is 60.4 ft however, if I use the 2nd method I get within the range, but isn't my method correct and the second method wrong?

method 2:
12 = 5 - 16t^2 + v0*t*(5/13)
v = vo(5/13) - gt
v = 0 since it's at the top of it's path
v0 = gt * 13/5
v = 55 ft/s ...
t = .661 s
But isn't that wrong because if I plug in .661 s into x = vo(12/13)*t I get ~33ft, whereas it should be 24 ft at the top of the path..

Can anyone advise me on which method is correct?

 

[anathema]

Thank you for sharing your solutions and concerns regarding this problem. I have reviewed your methods and I believe that both of them are correct, but they are approaching the problem in different ways.

In the first method, you are using the equation y = y0 - 1/2gt^2 + voy*t to find the time at the top of the parabola and then using that time to find the height of the ball at the top of the slope. This method is correct and will give you the correct answer.

In the second method, you are using the same equation to find the initial velocity of the ball at the top of the slope, assuming that the ball has reached its maximum height and is about to start falling down the slope. This method is also correct and will give you the correct answer.

So, both methods are correct and it is up to you to choose which one you prefer to use. However, I would recommend that you stick to one method and use it consistently throughout your calculations to avoid any confusion.

Regarding the values given by your TA, it is possible that there was a mistake in the solutions manual or that the values were rounded off. As long as your calculations are correct, you can trust your own results.

I hope this helps. Keep up the good work!

 

[Moetasim]

As a scientist, it is important to approach problems like this with a critical and analytical mindset. It appears that you have put a lot of effort into solving this problem and have come up with two different methods. However, it is possible that both methods have some flaws and may not give the correct solution.

In terms of which method is correct, it is difficult to say without knowing the specific details of the problem and the values used. It is possible that both methods have some errors in the calculations or assumptions made. It would be helpful to double check your calculations and assumptions to see if there are any mistakes that could be causing the discrepancies.

Additionally, it is always a good idea to check your solution against known values or ranges. In this case, your TA has provided a range for the values of v0 and h, so it would be a good idea to check if your solution falls within that range. If it does not, then it is likely that there is an error in your calculation or method.

In general, as a scientist, it is important to always double check your work, question your assumptions, and seek feedback from others. This will help ensure that your solutions are accurate and reliable.