Solving Kinematics in a Plane: Calculating Fred's Throw Speed

In summary: Thanks for the help and for the book reference.In summary, the conversation involves a question about kinematics and physics in a plane, specifically regarding a quarterback throwing a pass to a tight end. The problem involves using two fundamental kinematics equations to solve for the initial speed of the football thrown by the quarterback. After some initial confusion with the notation, it is determined that the equations need to be modified to account for the initial angle given. After correcting a typo, the correct solution is found with the help of others. The conversation concludes with the student expressing their gratitude for the help and acknowledging that they learned more by working through the problem themselves.
  • #1
geoffjb
165
1
I have a question involving kinematics and physics in a plane. The question is as follows:

Quarterback Fred is going to throw a pass to tight end Doug. Doug is 20 m in front of Fred and running straight away at 6.0 m/s when Fred throws the 500 g football at a 40 degree angle. Doug catches the ball without having to alter his speed and runs for the game-winning touchdown.

How fast did Fred throw the ball?

I am working with the two fundamental kinematics equations:

Code:
x = v_ix t

y = v_iy t - 1/2 gt^2

(An underscore, _, indicates subscripts. A caret, ^, indicates superscripts.)

Since the final positions of both the receiver and the ball must be identical, I have tried endlessly to modify equations which both involve position (that is, isolating x and then making one formula equal to another). However, some other variable (usually time) always gets in the way of my solving the equation. Any hints which would set me on the right track are appreciated.

Thanks.
 
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  • #2
As with most of these problems, the confusion arises because the student does not develop a sufficient notation (or he develops an unclear notation, which is not the case here).

Relative to Fred's position, Doug has the position as a function of time:
[tex]x_{D}(t)=6t+20[/tex]
Again, relative to Fred, the ball has the coordinate positions:
[tex]x_{B}(t)=v_{x}t=v_{0}\cos(\theta_{0})t, y_{B}(t)=v_{y}t-\frac{gt^{2}}{2}=v_{0}\sin(\theta_{0})t-\frac{gt^{2}}{2}[/tex]
where [itex]v_{0}[/itex] is the initial speed you are to find, and [itex]\theta_{0}[/itex] is the initial angle (given as 40 degrees).

Thus, essentially, you are to solve the system of equations for [itex]v_{0}[/itex], the other unknown being the time t:
[tex]6t+20=v_{0}\sin(\theta_{0})t[/tex]
[tex]v_{0}\sin(\theta_{0})t-\frac{gt^{2}}{2}=0[/tex]
knowing that [itex]t\neq{0}, v_{0}>0[/itex]
Agreed?
 
Last edited:
  • #3
arildno said:
As with most of these problems, the confusion arises because the student does not develop a sufficient notation (or he develops an unclear notation, which is not the case here).

Relative to Fred's position, Doug has the position as a function of time:
[tex]x_{D}(t)=6t+20[/tex]
Again, relative to Fred, the ball has the coordinate positions:
[tex]x_{B}(t)=v_{x}t=v_{0}\cos(\theta_{0})t, y_{B}(t)=v_{y}t-\frac{gt^{2}}{2}=v_{0}\sin(\theta_{0})t-\frac{gt^{2}}{2}[/tex]
where [itex]v_{0}[/itex] is the initial speed you are to find, and [itex]\theta_{0}[/itex] is the initial angle (given as 40 degrees).

Thus, essentially, you are to solve the system of equations for [itex]v_{0}[/itex], the other unknown being the time t:
[tex]6t+20=v_{0}\sin(\theta_{0})t[/tex]
[tex]v_{0}\sin(\theta_{0})t-\frac{gt^{2}}{2}=0[/tex]
knowing that [itex]t\neq{0}, v_{0}>0[/itex]
Agreed?

First of all, thank you arildno for your response.

Everything you said looks good. Here's what I did:

Since:

[tex]6t+20=v_{0}\sin(\theta)t[/tex]

and

[tex]v_{0}\sin(\theta)t=\frac{gt^{2}}{2}[/tex]

I calculated [tex]6t+20=\frac{gt^{2}}{2}[/tex]

From the quadratic formula, I got roots of -1.50 and 2.72 (which seem reasonable, considering the expected parabola of the projectile).

Using t = 2.72 s, I plugged the obtained value of t into the equation:

[tex]y=0=v_{0}\sin(\theta)t-\frac{gt^{2}}{2}[/tex]

So:

[tex]v_{0}\sin(\theta)t=\frac{gt^{2}}{2}[/tex]

[tex]v_{0}\sin(\theta)=\frac{gt}{2}[/tex]

[tex]v_{0}=\frac{gt}{2\sin(\theta)}[/tex]

Plugging everything in, I get a value for [itex]v_{0}[/itex] of 20.76.

When I submit this value, it claims it is incorrect. Does anyone see any errors along the way?
 
  • #4
Well, it looks like he made a typo:

[tex]6t+20=v_{0}\sin(\theta_{0})t[/tex] should actually be [tex]6t+20=v_{0}\cos(\theta_{0})t[/tex]
 
  • #5
Office_Shredder said:
Well, it looks like he made a typo:

[tex]6t+20=v_{0}\sin(\theta_{0})t[/tex] should actually be [tex]6t+20=v_{0}\cos(\theta_{0})t[/tex]

Ahhh, so it should.

So now I have:

[tex]6t+20=v_{0}\cos(\theta)t[/tex]

and

[tex]v_{0}\sin(\theta)t=\frac{gt^{2}}{2}[/tex]

So:

[tex]v_{0}=\frac{6t+20}{\cos(\theta)t}=\frac{gt}{2\sin(\theta)}[/tex]

I'll try this and see what happens.

Thanks, Shredder.
 
  • #6
This time, the quadratic equation yielded roots of -1.41 and 2.43.

Plugging t = 2.43 s into:

[tex]y=0=v_{0}\sin(\theta)t-\frac{gt^{2}}{2}[/tex]

or, as above:

[tex]v_{0}=\frac{gt}{2\sin(\theta)}[/tex]

Gives [itex]v_{0}=18.5 m/s[/itex]

Which is (finally) correct. Thanks to everyone who helped.
 
  • #7
Arrgh, grumph..yet another embarassment. Sorry about the typo in the system of equations at the end of my post..:redface:
 
  • #8
arildno said:
Arrgh, grumph..yet another embarassment. Sorry about the typo in the system of equations at the end of my post..:redface:

To be honest, I learned more by working through the problem after your suggested solution than I would have learned had the answer been correct.
 

Related to Solving Kinematics in a Plane: Calculating Fred's Throw Speed

What is kinematics in a plane?

Kinematics in a plane is the branch of physics that studies the motion of objects in a two-dimensional space. This includes analyzing the position, velocity, and acceleration of an object in a plane.

How do you calculate an object's speed in kinematics?

To calculate an object's speed in kinematics, you need to measure the distance traveled by the object and the time it took to travel that distance. Then, divide the distance by the time to get the average speed. This can be expressed as speed = distance/time.

What is Fred's throw speed?

Fred's throw speed refers to the velocity at which Fred throws an object in a two-dimensional plane. This can be calculated by measuring the distance the object travels and the time it takes to travel that distance.

What is the formula for calculating speed in kinematics?

The formula for calculating speed in kinematics is speed = distance/time. This is also known as the average speed formula.

Can you calculate an object's speed without knowing the distance?

No, it is not possible to calculate an object's speed without knowing the distance it traveled. Distance is an essential component in the speed formula and is necessary to determine an object's speed accurately.

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