# Projectile Motion (Ball thrown towards a wall)

## Homework Statement

In the figure, you throw a ball toward a wall at speed 34.0 m/s and at angle θ0 = 38.0˚ above the horizontal. The wall is distance d = 20.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall?

## Homework Equations

Projectile Motion Equations.

## The Attempt at a Solution

So I figured out A by using the trajectory equation (solved for y).
That value became my (y-yo) in the projectile versions of the kinematics equations.
Used that to solve for the final velocity, which came out to 13.6m/s

So I used 13.6m/s as my Vy, then found the components: [13.6cos(90),13.6sin(90)]

I got the answer for part A and for part C no problem.
For part B I got zero, because the cosine of 90 is zero. Wileyplus gave me trouble. Is there some special way to express a vector component with a magnitude of zero? Did I screw up here?

TSny
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So I used 13.6m/s as my Vy, then found the components: [13.6cos(90),13.6sin(90)]
Hello, Fetch. Not sure what you're doing in this step. What is the physical interpretation of using an angle of 90 degrees?

You already know the y component of velocity. You just need to find the x component.

Something to think about: What is the x component of the acceleration of the ball during the flight? What does that tell you about the x component of velocity during the flight?

Once I found the height (12.8952m) with the equation y=tan(theta)o*x-(gx^2/2[(Vocos(theta)o)^2]
I plugged that into the projectile equation:

Vo^2 = (Vo*sin(theta)o)^2-2g(y-yo)
Vo^2 = [(34.0m/s)*sin(38.0)]^2-2(9.8m/s)(12.8952)

At the end of all this I got a final velocity of 13.6m/s
Guess I assumed it was facing upwards at the time it hit the wall so I used 90 degrees as my angle when finding the components. In which case I found that 13.6m as the y component by luck. What's going on conceptually here?

TSny
Homework Helper
Gold Member
Once I found the height (12.8952m) with the equation y=tan(theta)o*x-(gx^2/2[(Vocos(theta)o)^2]
I plugged that into the projectile equation:

Vo^2 = (Vo*sin(theta)o)^2-2g(y-yo)
Why do you have a subscript "o" on the left side of the equation?

On the right side of the equation you are dealing with the y-component of the motion. [Note the occurrence of g, (y-yo) and sin(theta)]. So, the left side of the equation should also correspond to a y component.

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The subscript is just meant to indicate that its initial.

Are you saying I can shift this to correspond to the x component?

Okay.
So a few things happened this morning.

1) I solved the problem.
2) I took the time it took for the ball to make contact, the acceleration of 0 (horizontal motion in projectile equations), and the displacement. I got something like 26.809 using the kinematics equations for one dimensional motion and constant acceleration.

Anways, I used that as a vector then added my vectors together.

(13.6cos,13.6sin) + (26.8cos,26.8sin) = my components of the velocity when it made contact with the wall.

I have an idea of what happened here, but I don't understand completely. So the motions have no effect on the other, does that mean when calculating the components of vectors for these problems I have to do this all the time? My only thought on the matter is that its a matter of the object moving in two dimensional motion as opposed to one. Which is cool with me, but I want to make sure I'm right here. I have a bad habit of coming up with my own reasons why things worked then making them laws for myself. XD

TSny
Homework Helper
Gold Member
The velocity of the ball never makes a 90o angle to the horizontal. Also, I don't see why you are assuming a 0o angle in your calculations.

Anways, I used that as a vector then added my vectors together.

(13.6cos,13.6sin) + (26.8cos,26.8sin) = my components of the velocity when it made contact with the wall.
Why are you are adding together those two vectors? What does the second vector represent?

You are right that the horizontal motion should be treated separately from the vertical motion.

See if you can find the time it takes the ball to reach the wall by considering the horizontal component of the motion. What equation (in symbolic form) can you use to find the time?

I guess I just had it in my mind that to describe the projectile motion I needed to make calculations for strictly vertical (left, right) and horizontal (up, down) motions. The angles came as a result of moving counter clockwise from the positive x-axis.

First semester of Physics so I don't really have a solid foundation going on here.):

I'll get back to you later today with what I calculate for the time, I have a course coming up and need to shower and such. Assuming I don't find a tutor at my school to explain what's going on by that time. Thank you.

So it was by luck that I found those two answers as my components? Did I at least to part A correctly?

TSny
Homework Helper
Gold Member
I guess I just had it in my mind that to describe the projectile motion I needed to make calculations for strictly vertical (left, right) and horizontal (up, down) motions. The angles came as a result of moving counter clockwise from the positive x-axis.
OK. I think I see now how you are constructing the expressions with the 0 and 90 degree angles. You are treating the x and y components of velocity as vectors and then adding them to get the velocity vector for the ball. Your result is correct for the velocity as the ball hits the wall.

Normally, when you are asked for the x and y component of a velocity, you would just specify numbers (not vectors) for vx and vy. So, for example, the answer to (b) would just be vx = 26.8 m/s.

I get the same answer as you for part (a). So, you probably worked it correctly although you didn't show any details of your calculation other than specify the equation you used.