Projectile motion with people moving

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Homework Help Overview

The problem involves projectile motion, specifically a scenario where a quarterback throws a football to a running tight end. The tight end is moving away from the quarterback at a constant speed, and the problem requires determining the initial speed of the football thrown at an angle.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations of motion related to projectile dynamics and attempt to express the initial velocity in terms of time and other variables. There are questions about solving for time and ensuring the equations are set up correctly.

Discussion Status

Participants are actively engaging with the problem, sharing their attempts to manipulate equations and solve for variables. Some guidance has been offered regarding the quadratic nature of the equations and the need to check which equation to use for final calculations. There is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on ensuring the equations are correctly set up and interpreted.

sunbunny
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I'm having trouble with this problem as well:

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 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 don't even know where to start with so so if anyone can help me get started that would be great.

Thanks
 
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Look at these two equations and figure out how and why they work:

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

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

Solve each in terms of v_{0}, equate one equation to the other, solve for t, and then plug it into a formula to find v_{0}.
 
Okay, so when i solved for vo i for:

vo= 6t+20/cost and vo= gt^2/2sint

then i made these equations equal to each other:


6t + 20/cost = gt^2/sint

but when i solve for t, the value that i get donesn't work. if you could explain how to solve for t properly that would be geat.

Thank you for your response
 
sunbunny said:
Okay, so when i solved for vo i for:

vo= 6t+20/cost and vo= gt^2/2sint

then i made these equations equal to each other:6t + 20/cost = gt^2/sint

but when i solve for t, the value that i get donesn't work. if you could explain how to solve for t properly that would be geat.

Thank you for your response

The equation you end up with is a quadratic. Manipulate it to form ax^{2}+bx+c=0, and solve for x. You will get two roots-only one will be applicable.
 
Last edited:
okay this time when i did that i got:

12sint^2 + 40sint=gcost^3

-gcost^3+ 12sint^2 + 40sint=0

t(-gcost^2 + 12sint =40sin)=0

i solved for inside the brackets using the quadratic formula and i got 2.43 for my time and then i put it back into one of the equations and i got 78m/s however i still got the wrong answer.

I was wondering if you could please check over my work to make sure that I'm not missing anything. Thanks!
 
sunbunny said:
okay this time when i did that i got:

12sint^2 + 40sint=gcost^3

-gcost^3+ 12sint^2 + 40sint=0

t(-gcost^2 + 12sint =40sin)=0

i solved for inside the brackets using the quadratic formula and i got 2.43 for my time and then i put it back into one of the equations and i got 78m/s however i still got the wrong answer.

I was wondering if you could please check over my work to make sure that I'm not missing anything. Thanks!

2.43 s is the correct time. Which equation did you plug it back into?
 
i put it into the 6t+20/cost equation
 
sunbunny said:
i put it into the 6t+20/cost equation

Put it into:

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

Remember, at the time you've specified, y will have a certain value.

Your question is solved with this equation.
 
okay i got it now, thank you for your help!
 

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