2 Two-dimensional motion problems

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In summary, if you want to know how far a ball will travel after it has been thrown and has bounced off the ground, you need to solve for the initial velocity and the time of flight.
  • #1
MJay82
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I spent all day yesterday trying to figure out how to do these on my own, but I can't seem to conceptualize these two problems.

Problem 1

Homework Statement


Background
When baseball outfielders throw the ball, they usually allow it to take one bounce on the theory that the ball arrives sooner this way. Suppose that after the bounce the ball rebounds at the same angle θ as it had when released (see figure below) but loses half its speed.

Problem
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 as one thrown upward at ϕ = 43.8° with no bounce?

Homework Equations


My teacher said I needed to use this trig identity:
sin(2u) = 2sin(u)cos(u)

The Attempt at a Solution


I attemped this problem in this manner.
R(1bounce) = (Vi sin2R2R(1 bounce) = R(0 bounce)
sin(2[tex]\Phi[/tex])=2sin([tex]\Theta[/tex])cos([tex]\Theta[/tex])
.4996=sin([tex]\Theta[/tex])cos([tex]\Theta[/tex])This doesn't seem right. Or at least I don't know the trig identity that can get me [tex]\Theta[/tex] from here.For this second problem, I figured out the initial height and the range, but I don't know how to work these into a problem that can get the time I need at part C.

Homework Statement


A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.50 m/s at an angle of 16.0° below the horizontal. It strikes the ground 6.00 s later.

(a) How far horizontally from the base of the building does the ball strike the ground? (Xf)
(b) Find the height from which the ball was thrown. (Yi)
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Xi = 0 m, Xf = range = 43.26 m
Yi = 188 m, Yi = 0 m
Vxi = 7.21 m/s, Vyi = -2.07 m/s
Vxf = 7.21 m/s, Vyf = -2.07 m/s - 9.8t
Ax = 0 m/s^2, Ay = -9.8 m/s^2
t = 6.00 s

Homework Equations


Kinematic equations

The Attempt at a Solution


I initially thought this part (part c) was easy - I just used Yf = Yi + (Vyi)t + (a/2)t^2.

Obviously I didn't account for the change in displacement of X, and treated the problem as if the object was dropped straight down. So I'm stuck, and don't know where to go from here. Any help is greatly appreciated.
 
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  • #2
Problem 1 is most interesting; thanks for posting it.
I don't see how you did it. Surely you need to work out the vertical part of the motion to get the time of flight? Then the horizontal distance of the flight. I don't see any time or distance calcs in your work.

You can solve .4996 = sin(θ)*cos(θ) by using that identity:
2*.4996 = 2sin(θ)*cos(θ)
2*.4996 = sin(2θ)
I see that gives θ = 43.8, but of course that is not the answer to the question. You would have to throw at a smaller angle so the ball hits the ground after a smaller distance, leaving room for the bounce.
 
  • #3
Thanks for the reply, Delphi. This assignment was due last night. I actually ended up getting the first question correct by creating a table of answers to problems similar to this one. Not the way I would have liked to solve it, but I'll take a 90 over a 70 and learn how to do it next week. The answer was 26.5 degrees.
 
  • #4
For next week . . . nearly all projectile motion questions yield to this elegant method:
- make headings for horizontal and vertical parts of the motion
(the two parts are independent; one does not affect the other)
- there are no horizontal forces, the motion is not accelerated so you write d = vt or perhaps x = Vx*t if you are picky.
- under the vertical heading, you write d = Vi*t + ½at² and
v = Vi + at (though this last formula is not needed in this particular question)
- fill in the things you know. For the horizontal part, I wrote
147 = v*cos(37)*t (where v is the initial speed of the ball).
This gives t = 147/(v*cos(37)). [1]
In the vertical part, you get
26 = v*sin(37)*t - ½gt² [2]
The classic two equations with two unknowns. Sub 1 into 2 and you soon get a number for the initial velocity. And the time of flight if you want it.

The next step is to use that initial velocity for the throw with a bounce. This time you use the velocity equation for the vertical part and not the distance one, just because the vertical distance is neither known nor needed.

Don't let this one go by . . . it is a very widely needed technique!
 
  • #5


Hello,

I understand that two-dimensional motion problems can be challenging to conceptualize and solve. It seems like you have made some good attempts at these problems, but there are a few areas where you may have gone wrong.

In the first problem, you correctly identified the trig identity that needs to be used, but your equation is not entirely correct. The correct equation should be sin(2ϕ) = 2sin(ϕ)cos(ϕ). Also, you have used the wrong variable for the angle θ. It should be θ in both the sin and cos terms. Once you have the correct equation, you can solve for θ by setting it equal to the given angle of ϕ = 43.8° and solving for θ. This will give you the angle at which the ball should be thrown to go the same distance with one bounce as it would with no bounce.

For the second problem, you have correctly identified the initial height and range, but your equations for the initial and final velocities are not entirely accurate. The initial velocity should be broken down into its x and y components, which would be 7.21 m/s for Vxi and -2.07 m/s for Vyi. The final velocity in the y-direction should be calculated using the equation Vyf = Vyi + at, where a is the acceleration due to gravity, -9.8 m/s^2. Once you have these values, you can use the kinematic equations to solve for the remaining variables. For part c, you can use the equation Yf = Yi + Vyi*t + (1/2)a*t^2, but you will also need to take into account the horizontal displacement, which can be calculated using the equation X = Vxi*t. Once you have both the vertical and horizontal displacements, you can use the Pythagorean theorem to find the total displacement and solve for the time t.

I hope this helps you in solving these two-dimensional motion problems. Remember to always break down the initial velocity into its components and consider both the vertical and horizontal motions separately. Good luck!
 

Related to 2 Two-dimensional motion problems

1. What is two-dimensional motion?

Two-dimensional motion refers to the movement of an object in two directions, typically referred to as the x-axis and y-axis. This type of motion is commonly seen in everyday life, such as a ball being thrown in the air or a person walking in a straight line.

2. How is two-dimensional motion different from one-dimensional motion?

One-dimensional motion only involves movement in one direction, while two-dimensional motion involves movement in two directions. This means that two-dimensional motion problems require the consideration of both horizontal and vertical components, while one-dimensional motion problems only involve one component.

3. What are some common equations used to solve two-dimensional motion problems?

The most commonly used equations for two-dimensional motion are the equations for displacement, velocity, and acceleration, which are derived from the basic principles of kinematics. These equations include the Pythagorean theorem for finding the magnitude of the displacement, as well as equations for finding the x and y components of velocity and acceleration.

4. How can vectors be used to represent two-dimensional motion?

Vectors are commonly used to represent two-dimensional motion because they can include both magnitude (size or speed) and direction. In two-dimensional motion, vectors are often broken down into their horizontal and vertical components to accurately represent the motion of an object in both directions.

5. How can two-dimensional motion problems be applied in real-life situations?

Two-dimensional motion problems can be applied in various real-life situations, such as calculating the trajectory of a projectile, determining the velocity and acceleration of a moving object, or predicting the motion of an object in a roller coaster or amusement park ride. Understanding two-dimensional motion can also help in fields such as engineering, physics, and sports.

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