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

[geoffjb]

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:

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.

 

[arildno]

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:
x_{D}(t)=6t+20
Again, relative to Fred, the ball has the coordinate positions:
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}
where v_{0} is the initial speed you are to find, and \theta_{0} is the initial angle (given as 40 degrees).

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

 

[geoffjb]

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:
x_{D}(t)=6t+20
Again, relative to Fred, the ball has the coordinate positions:
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}
where v_{0} is the initial speed you are to find, and \theta_{0} is the initial angle (given as 40 degrees).

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

First of all, thank you arildno for your response.

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

Since:

6t+20=v_{0}\sin(\theta)t

and

v_{0}\sin(\theta)t=\frac{gt^{2}}{2}

I calculated 6t+20=\frac{gt^{2}}{2}

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:

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

So:

v_{0}\sin(\theta)t=\frac{gt^{2}}{2}

v_{0}\sin(\theta)=\frac{gt}{2}

v_{0}=\frac{gt}{2\sin(\theta)}

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

When I submit this value, it claims it is incorrect. Does anyone see any errors along the way?

 

[Office_Shredder]

Well, it looks like he made a typo:

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

 

[geoffjb]

Well, it looks like he made a typo:

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

Ahhh, so it should.

So now I have:

6t+20=v_{0}\cos(\theta)t

and

v_{0}\sin(\theta)t=\frac{gt^{2}}{2}

So:

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

I'll try this and see what happens.

Thanks, Shredder.

 

[geoffjb]

This time, the quadratic equation yielded roots of -1.41 and 2.43.

Plugging t = 2.43 s into:

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

or, as above:

v_{0}=\frac{gt}{2\sin(\theta)}

Gives v_{0}=18.5 m/s

Which is (finally) correct. Thanks to everyone who helped.

 

[arildno]

Arrgh, grumph..yet another embarassment. Sorry about the typo in the system of equations at the end of my post..

 

[geoffjb]

Arrgh, grumph..yet another embarassment. Sorry about the typo in the system of equations at the end of my post..

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.