Speed at impact of 2 balls thrown from a building

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Homework Help Overview

The discussion revolves around a physics problem involving two balls thrown from a height of 7.25 meters, one horizontally and the other straight down, both with an initial speed of 63.5 mph. Participants are exploring the calculation of their impact speeds and the difference between them, referencing a textbook answer of -9.52 m/s.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the equation v_{y}^2 = v_{y0}^2 - 2g(y - y_{0}) and question its application, particularly regarding the signs and values for initial and final positions. There is also mention of converting speeds from mph to m/s and the implications of air resistance being negligible.

Discussion Status

Some participants have provided guidance on correcting the positions in the equation, which has led to revised calculations. However, there remains uncertainty as the calculated differences in speeds do not align with the textbook answer, prompting further inquiry into potential errors in reasoning or assumptions.

Contextual Notes

Participants note that the problem is considered basic and that air resistance is not a factor. There is an ongoing exploration of the correct interpretation of the variables involved in the equations used.

KarlKarlJohn
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Hi, I'm having a bit of trouble with the below practice problem from my textbook. The answer given in the back is -9.52 m/s, but there is no explanation. Thanks for any help.

Homework Statement


Two balls are thrown off of a 7.25 m high building both at an initial speed of 63.5 mph. However, one ball is thrown horizontally and the other straight down. What is the difference in the speed of the balls when they hit the ground. This is a basic problem, so things such as air resistance don't need to be taken into account.


Homework Equations


I believe that v_{y}^2 = v_{y0}^2 - 2g(y - y_{0}) is relevant here. However, that equation was introduced a few chapters before the chapter this problem is found in, so maybe I am supposed to use a different method involving Kinetic Energy, Work, or Power which are the subjects of the current chapter.


The Attempt at a Solution


I converted the initial speed from mph into m/s and got 28.38 m/s. I then tried to solve for v_{y}^2 in the above noted equation for each ball. For ball 1 (thrown horizontally) I used:
v_{y}^2 = 0^2 - 2(9.81)(7.25-0) My first problem here is that this gives me a negative sign on the right side of the equation, so I can't take the square root of both sides to solve for v. I did see an example in the book though where the negative sign mysteriously disappeared, so assuming this v = \sqrt{2*9.81*7.25} = 11.93m/s
For ball 2 (thrown straight down) I solved for v with the same equation:
v_{y}^2 = 28.38^2 - 2(9.81)(7.25-0) and got v = 25.75 m/s. The difference in these values in 13.82 m/s, which is incorrect as they should differ by 9.52 m/s according to the book.
 
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KarlKarlJohn said:
Hi, I'm having a bit of trouble with the below practice problem from my textbook. The answer given in the back is -9.52 m/s, but there is no explanation. Thanks for any help.

Homework Statement


Two balls are thrown off of a 7.25 m high building both at an initial speed of 63.5 mph. However, one ball is thrown horizontally and the other straight down. What is the difference in the speed of the balls when they hit the ground. This is a basic problem, so things such as air resistance don't need to be taken into account.

Homework Equations


I believe that v_{y}^2 = v_{y0}^2 - 2g(y - y_{0}) is relevant here. However, that equation was introduced a few chapters before the chapter this problem is found in, so maybe I am supposed to use a different method involving Kinetic Energy, Work, or Power which are the subjects of the current chapter.

The Attempt at a Solution


I converted the initial speed from mph into m/s and got 28.38 m/s. I then tried to solve for v_{y}^2 in the above noted equation for each ball. For ball 1 (thrown horizontally) I used:
v_{y}^2 = 0^2 - 2(9.81)(7.25-0) My first problem here is that this gives me a negative sign on the right side of the equation, so I can't take the square root of both sides to solve for v. I did see an example in the book though where the negative sign mysteriously disappeared, so assuming this v = \sqrt{2*9.81*7.25} = 11.93m/s
For ball 2 (thrown straight down) I solved for v with the same equation:
v_{y}^2 = 28.38^2 - 2(9.81)(7.25-0) and got v = 25.75 m/s. The difference in these values in 13.82 m/s, which is incorrect as they should differ by 9.52 m/s according to the book.
Hello KarlKarlJohn. Welcome to PF !

For ball 2:

y = 0 ,

y0 = 7.25 .

You have them reversed, so you get a negative.
 
SammyS said:
Hello KarlKarlJohn. Welcome to PF !

For ball 2:

y = 0 ,

y0 = 7.25 .

You have them reversed, so you get a negative.

Thanks for the help and the welcome! Switching those values makes sense and definitely helps with my issue with the negative, but I still don't seem to be getting the correct answer. I now have: v_{y}^2 = 0^2 - 2(9.81)(-7.25) \Rightarrow v = 11.92m/s
v_{y}^2 = 28.38^2 - 2(9.81)(-7.25) \Rightarrow v = 30.78 m/s

The difference in these velocities is 18.86 m/s, my book says the answer should be -9.52 m/s. Any idea what else I may be doing wrong?
 
What about horizontal speed?
 

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