# Projectile Motion problem Phyisics 1

• cbchapm2
In summary: I think you can still make some assumptions about the height of the bounce compared to the height of the thrown ball, and use some ratios...In summary, this conversation is about a physics problem involving a baseball outfielder throwing a ball and the trajectory of the ball after it bounces. The problem asks for the initial angle at which the ball should be thrown in order for it to travel the same distance with one bounce as it would with no bounce, as well as the ratio of the times for the one-bounce and no-bounce throws. The conversation includes a discussion of kinematic equations and the assumption of no friction and perfectly elastic bounces.
cbchapm2
1. Homework Statement
When baseball outfielders throw the ball, they usually allow it to take one bounce, on the theory that the ball arrives at its target sooner that way. Suppose that, after the bounce, the ball rebounds at the same angle θ that it had when it was released (as in the figure below), but loses half its speed.

ETA: Hopefully this figure shows up...I attached it.

(a) Assuming the ball is always thrown with the same initial speed, at what angle θ should the ball be thrown in order to go the same distance D with one bounce (blue path) as a ball thrown upward at phi = 29.2° with no bounce (green path)?
(b) Determine the ratio of the times for the one-bounce and no-bounce throws.

2. Homework Equations
(a) I really don't know where to start.

(b) t1b/t0b

3. The Attempt at a Solution

I have no clue where to start. This is my first physics class and I'm stuck with an awful professor, and I'm lost. Help?

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Where is the figure?

Added it!

You must have studied uniformly accelerated motion in order to solve this. What are its principal equations?

Uhm, like I said I said I really don't know where to start. I know my kinematic equations and I have tried to make sense of the problem using my book (Serway Vuille, 9th edition) but I really just need help understanding this.

You should start by writing down kinematic equations for 2D motion. The Y-direction (height) has constant acceleration g (gravity) acting downward, the X-direction has no acceleration. What equations govern the flight of the ball after it was launched and before it touches down?

I really have no clue.

I know V0x=V0cos(theta) and V0y=V0sin(theta), but I have no V0 so how do I get anything from that? All I have is the angle from one of the throws.

Also, why does the X direction have no acceleration?

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cbchapm2 said:
I really have no clue.

I know V0x=V0cos(theta) and V0y=V0sin(theta), but I have no V0 so how do I get anything from that? All I have is the angle from one of the throws.

Also, why does the X direction have no acceleration?

Once the ball is thrown, it will have a constant horizontal velocity (ignoring air resistance), but in the vertical, the acceleration of gravity exerts a force downward on the ball. That's why the trajectory follows a parabolic arc.

What is the equation for distance as a function of time in the horizontal (constant velocity)?

What is the equation for height as a function of time in the vertical (where there is an initial Voy and the acceleration of gravity points downward...? Those are the standard kinematic equations that the other helper was trying to get you to write down for us.

berkeman said:
Once the ball is thrown, it will have a constant horizontal velocity (ignoring air resistance), but in the vertical, the acceleration of gravity exerts a force downward on the ball. That's why the trajectory follows a parabolic arc.

What is the equation for distance as a function of time in the horizontal (constant velocity)?

What is the equation for height as a function of time in the vertical (where there is an initial Voy and the acceleration of gravity points downward...? Those are the standard kinematic equations that the other helper was trying to get you to write down for us.

distance as function of time: deltaX=V0t + 1/2at^2, is that what you're looking for? So I have no t, I have no V0, I have no deltaX, and a= -9.8...

Here's the rest of my kinematic equations, happy berkeman?
V=V0 + at
2a(deltaX)=V^2-V0^2
deltaX=1/2(V + V0)t

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cbchapm2 said:
distance as function of time: deltaX=V0t + 1/2at^2, is that what you're looking for? So I have no t, I have no V0, I have no deltaX, and a= -9.8...

There is no acceleration in the x direction, right? But there is in the y direction, as you indicate it is -9.8m/s^2.

What are the equations in the x & y directions for the position as a function of time of the thrown ball? Don't worry that you aren't given numbers for things like Vo and so on -- the question is asking for angles and relative times. Just use variables for now on all of those quantities and look for them to ratio out in the end results...

berkeman said:
There is no acceleration in the x direction, right? But there is in the y direction, as you indicate it is -9.8m/s^2.

What are the equations in the x & y directions for the position as a function of time of the thrown ball? Don't worry that you aren't given numbers for things like Vo and so on -- the question is asking for angles and relative times. Just use variables for now on all of those quantities and look for them to ratio out in the end results...

So you just disregard a in the x-direction? I still don't understand why there isn't any at all. Would you use the equation: deltaX=1/2(V + V0)t since it doesn't have any acceleration in it at all? If there's no numbers, how do you find the other angle?

What kind of friction are you assuming? What kind of elasticity are you assuming for the bounce? If there is no friction in the air or at the bounce, and if the bounce is 'perfectly elastic", the bounce would be exactly half way between the fielder and the catcher. You must either be given an initial speed or use a "placekeeper", like "v" and then set it to give the right distance.

cbchapm2 said:
So you just disregard a in the x-direction? I still don't understand why there isn't any at all. Would you use the equation: deltaX=1/2(V + V0)t since it doesn't have any acceleration in it at all? If there's no numbers, how do you find the other angle?

This problem is made a bit harder in the x-direction because they say the ball loses half of its speed at the bounce. So yes, there is a discontinuity there that you could call a deceleration in the x (and y) directions. But only at the bounce for the x-direction.

Normally you work projectile motion problems by using the constant Vx to give you an equation relating time and distance, and then use the vertical motion equation (including the acceleration down from gravity) to give you a 2nd equation relating time and vertical position. At the end of the trajectory, the x and y values have to come together at the end point, and you use that to help you solve the equations.

In this problem, for the bounce trajectory, you have to use the fact that y=0 at the first bounce, then write the equations for the 2nd part of the travel based on the new velocity of Vo/2.

berkeman said:
This problem is made a bit harder in the x-direction because they say the ball loses half of its speed at the bounce. So yes, there is a discontinuity there that you could call a deceleration in the x (and y) directions. But only at the bounce for the x-direction.

Normally you work projectile motion problems by using the constant Vx to give you an equation relating time and distance, and then use the vertical motion equation (including the acceleration down from gravity) to give you a 2nd equation relating time and vertical position. At the end of the trajectory, the x and y values have to come together at the end point, and you use that to help you solve the equations.

In this problem, for the bounce trajectory, you have to use the fact that y=0 at the first bounce, then write the equations for the 2nd part of the travel based on the new velocity of Vo/2.

Look thanks for trying but none of this is making sense. I can't understand this and actually apply it.

Okay, will the instructor or TA go over it in class? Can you ask the TA about it in person?

Nope, my test is Wednesday night. We don't have TA's in this class, yeah I know, weird. My professor is terrible at explaining anything, hence why NOTHING makes sense to me.

cbchapm2 said:
Nope, my test is Wednesday night. We don't have TA's in this class, yeah I know, weird. My professor is terrible at explaining anything, hence why NOTHING makes sense to me.

Well, maybe try a practice exercise to see if it helps your understanding of how to use these types of equations.

Different problem -- a 45 degree launch angle for a projectile maximizes the range (if no air resistance). Find how much faster the projectile has to be launched at a 55 degree angle in order to achieve the same distance to impact. Express the answer as a ratio of initial velocities.

So if you can do this more basic practice problem, you will be well on your way to solving the bounce problem, and in being prepared and more comfortable for the upcoming test.

berkeman said:
Well, maybe try a practice exercise to see if it helps your understanding of how to use these types of equations.

Different problem -- a 45 degree launch angle for a projectile maximizes the range (if no air resistance). Find how much faster the projectile has to be launched at a 55 degree angle in order to achieve the same distance to impact. Express the answer as a ratio of initial velocities.

So if you can do this more basic practice problem, you will be well on your way to solving the bounce problem, and in being prepared and more comfortable for the upcoming test.

I seriously don't know how to even begin solving a problem like this. I'm not trying to make you do the problem for me, obviously, since this one is not a homework problem. I really don't know where to start. I need to be walked through the steps of an example problem before I really understand how to do it with other problems.

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cbchapm2 said:
I seriously don't know how to even begin solving a problem like this. I'm not trying to make you do the problem for me, obviously, since this one is not a homework problem. I really don't know where to start. I need to be walked through the steps of an example problem before I really understand how to do it with other problems.

This is a good review of the concepts. Give it a read (you'll see the equations that I've been talking about in there), and see if that helps you solve some basic projectile motion problems. You can also read some of the many other threads in this Intro Physics forum, to see how others are working through similar problems.

http://en.wikipedia.org/wiki/Projectile_motion

.

berkeman said:
This is a good review of the concepts. Give it a read (you'll see the equations that I've been talking about in there), and see if that helps you solve some basic projectile motion problems. You can also read some of the many other threads in this Intro Physics forum, to see how others are working through similar problems.

http://en.wikipedia.org/wiki/Projectile_motion

.

When I was attempting to do this on my own I found this particular page and was still unable to apply it.

Why don't you just walk me through the example problem you gave?

cbchapm2 said:
Why don't you just walk me through the example problem you gave?

I'm about to leave work, but I'll give you a start.

For the 45 degree case:

X(t) = Xi + (Vx * t)
Y(t) = Yi + (Viy * t) - 1/2(g * t^2)

Look familiar? Those are the basic equations you use for most of these projectile motion questions. Because there is no horizontal acceleration, there is no 3rd term on the right for the X1(t) equation.

For the 55 degree case, you have the same equations, just a difference in the x & y components of the intial velocity. In general, you have:

Viy = Vi * sin(theta)
Vix = Vi * cos(theta)

So now you should be able to write the two sets of equations -- one for theta=45 degrees, and one for theta=55 degrees. The only difference is the change in the components of the initial velocity Vi between the two cases.

Then to solve my practice question, you could just assume some Vi for the 45 degree case (say, 10m/s), then use the two position equations for the 45 degree case to find how far downrange the projectile lands. Then use that x distance to work back through the 55 degree case to find the higher Vi that is required to achieve that distance with the higher launch angle.

Give it a go, and show us your work here. We can correct any mistakes that you make, and help you get more comfortable working the kinematic equations to solve projectile motion problems.

EDIT -- I fixed a couple typos, so use this version of my post.

Jay Singh said:
this case is not possible because once the ball touches the ground it will lose PE energy and won't bounce at the same angle...so this ques aint right..

It's obviously a simplification to make the problem easier to solve. Do you not see that?

berkeman said:
I'm about to leave work, but I'll give you a start.

For the 45 degree case:

X(t) = Xi + (Vx * t)
Y(t) = Yi + (Viy * t) - 1/2(g * t^2)

Look familiar? Those are the basic equations you use for most of these projectile motion questions. Because there is no horizontal acceleration, there is no 3rd term on the right for the X1(t) equation.

For the 55 degree case, you have the same equations, just a difference in the x & y components of the intial velocity. In general, you have:

Viy = Vi * sin(theta)
Vix = Vi * cos(theta)

So now you should be able to write the two sets of equations -- one for theta=45 degrees, and one for theta=55 degrees. The only difference is the change in the components of the initial velocity Vi between the two cases.

Then to solve my practice question, you could just assume some Vi for the 45 degree case (say, 10m/s), then use the two position equations for the 45 degree case to find how far downrange the projectile lands. Then use that x distance to work back through the 55 degree case to find the higher Vi that is required to achieve that distance with the higher launch angle.

Give it a go, and show us your work here. We can correct any mistakes that you make, and help you get more comfortable working the kinematic equations to solve projectile motion problems.

EDIT -- I fixed a couple typos, so use this version of my post.

Okay, so I'm trying to find the Vx and Vy for 45*. I just used a value for Vi like you said, 10 m/s.
Vx=10m/s(cos 45*)=7.071
Vy=10m/s(sin 45*)=7.071

So now I plug those into the respective X(t) and Y(t) equations? What value do I use for the Xi and Yi of them?

cbchapm2 said:
Okay, so I'm trying to find the Vx and Vy for 45*. I just used a value for Vi like you said, 10 m/s.
Vx=10m/s(cos 45*)=7.071
Vy=10m/s(sin 45*)=7.071

So now I plug those into the respective X(t) and Y(t) equations? What value do I use for the Xi and Yi of them?

In this problem, assume the projectile starts at ground level and ends at ground level. So Yi is zero, and you can choose to make the start point the origin, so Xi is zero.

In more complicated projectile motion problems, the projectile may start and/or end at some non-zero height.

cbchapm2 said:
Okay, so I'm trying to find the Vx and Vy for 45*. I just used a value for Vi like you said, 10 m/s.
Vx=10m/s(cos 45*)=7.071
Vy=10m/s(sin 45*)=7.071
One thing that will help is to solve problems using symbols instead of numbers. That will let you (and others) see what is going on in the equations, see what the physics is. Plugging in numbers right away hides that.

## What is projectile motion in physics?

Projectile motion is the motion of an object thrown or launched into the air at an angle, under the force of gravity. It follows a curved path due to the combination of its initial horizontal velocity and vertical acceleration from gravity.

## What are the key factors that affect projectile motion?

The key factors that affect projectile motion are the initial velocity, launch angle, and the acceleration due to gravity. Air resistance and wind can also affect the trajectory of a projectile.

## How do you calculate the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the formula: R = (V02 sin 2θ)/g, where R is the range, V0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## What is the maximum height reached by a projectile?

The maximum height reached by a projectile is known as the apex or peak of the trajectory. It can be calculated using the formula: H = (V02 sin2 θ)/(2g), where H is the maximum height, V0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## How does air resistance affect projectile motion?

Air resistance can affect the trajectory of a projectile by slowing it down and altering its path. This is because air resistance acts in the opposite direction of the projectile's motion, reducing its horizontal and vertical velocities.

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