Speed at impact of 2 balls thrown from a building

[KarlKarlJohn]

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 $v = \sqrt{2*9.81*7.25} = 11.93$m/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.

 

[SammyS]

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 $v = \sqrt{2*9.81*7.25} = 11.93$m/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.

 

[KarlKarlJohn]

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?

 

[haruspex]

What about horizontal speed?

 

[Nickie]

Hello,

Thank you for reaching out for help with this problem. As a scientist, it is important to carefully consider the equations and assumptions being made in order to arrive at an accurate solution. Let's take a closer look at the problem and the steps you have taken so far.

First, let's define the variables that we will be using:
- v_y = final speed in the vertical direction (m/s)
- v_{y0} = initial speed in the vertical direction (m/s)
- g = acceleration due to gravity (9.81 m/s^2)
- y = final height (m)
- y_{0} = initial height (m)

Now, let's consider the two balls being thrown from the building. Ball 1 is thrown horizontally, meaning that its initial speed in the vertical direction (v_{y0}) is 0 m/s. Ball 2 is thrown straight down, meaning that its initial speed in the vertical direction is the same as its initial speed overall (v_{y0} = 28.38 m/s).

To solve for the final speed (v_y) of each ball, we can use the equation v_{y}^2 = v_{y0}^2 - 2g(y - y_{0}). However, it is important to note that this equation only applies to objects that are in free fall, meaning that the only force acting on the object is gravity. In this case, we are assuming that air resistance is not a factor, so this equation is appropriate to use.

For ball 1, we have:
v_{y}^2 = 0^2 - 2(9.81)(7.25-0)
v_{y}^2 = -141.675
This is where your confusion with the negative sign comes in. In this equation, the negative sign indicates that the final speed (v_y) is in the opposite direction of the initial speed (v_{y0}). However, we are interested in the magnitude of the final speed, so we can simply take the absolute value of v_y and get:
|v_y| = 11.93 m/s

For ball 2, we have:
v_{y}^2 = 28.38^2 - 2(9.81)(7.25-0)
v_{y}^2 = 522.984
Taking the square root of this value gives us the final