Projectile Motion of football receiver

[emilytm]

Homework Statement

A football receiver running straight downfield at 5.5m/s is 10m in front of the quarterback when a pass is thrown downfield at 25 degrees above the horizon. If the receiver never changes speed and the ball is caught at the same height from which it was thrown find initial velocity of the football, the amount of time it spends in the air, and the distance between the quarterback and receiver when the catch is made.

Homework Equations

The Attempt at a Solution

I drew a diagram, found Vx=Vo*cos theta and Vy=Vo*sin theta. I also came up with the equations t=Vo*sin theta/ 9.81m/s^2 and t/2=delta x/ Vo*cos theta. Am I on the right track?

 

[TSny]

I drew a diagram

Good

I found Vx=Vo*cos theta and Vy=Vo*sin theta.

OK [Edit: As @berkeman points out below, the second equation is only correct if Vy represent the y component of initial velocity, V_0y.]

I also came up with the equations t=Vo*sin theta/ 9.81m/s^2

What time does this represent? Can you explain how you arrived at this expression?

 

[berkeman]

Homework Statement

A football receiver running straight downfield at 5.5m/s is 10m in front of the quarterback when a pass is thrown downfield at 25 degrees above the horizon. If the receiver never changes speed and the ball is caught at the same height from which it was thrown find initial velocity of the football, the amount of time it spends in the air, and the distance between the quarterback and receiver when the catch is made.

Homework Equations

The Attempt at a Solution

I drew a diagram, found Vx=Vo*cos theta and Vy=Vo*sin theta. I also came up with the equations t=Vo*sin theta/ 9.81m/s^2 and t/2=delta x/ Vo*cos theta. Am I on the right track?

Welcome to the PF.

Can you post how you came up with the last parts? Vy=Vo*sin theta is not right because the vertical velocity as a function of time depends on the acceleration of gravity.

On my initial sketch, I've assumed no air resistance, so the angle of the ball being caught matches the launch angle. The y height matches the initial thrown height (that gives an equation in y), and the equation in x involves the speed and initial position of the receiver. Can you post those equations as part of your work?

 

[berkeman]

Vy=Vo*sin theta is not right

Oh, you probably meant to write Vyo=Vo*sin theta, not Vy(t) =Vo*sin theta. It's important to write the equations for x(t) and Vy(t) and y(t) as part of your solution...

EDIT / ADD -- Oops, I missed that TSny had posted...

 

[emilytm]

Good
OK [Edit: As @berkeman points out below, the second equation is only correct if Vy represent the y component of initial velocity, V_0y.]
What time does this represent? Can you explain how you arrived at this expression?

It represented the time when the football was at maximum height.

Good
OK [Edit: As @berkeman points out below, the second equation is only correct if Vy represent the y component of initial velocity, V_0y.]
What time does this represent? Can you explain how you arrived at this expression?

Yes I did mean to write Vyo.
I found time from the equation: delta x=Vo*cos theta*t.

 

[emilytm]

For the first equation for time, I used Vy=Vyo+ay*t at Vy=0 and Vyo=Vo*sin theta (time at max height which is half the total time).

For the second equation, I used delta x=Vo*cos theta*t for total time.

 

[TSny]

For the first equation for time, I used Vy=Vyo+ay*t at Vy=0 and Vyo=Vo*sin theta (time at max height which is half the total time).

OK, good.

For the second equation, I used delta x=Vo*cos theta*t for total time.

OK, but I'm a little confused with your notation in your first post. You used the symbol t for the time to reach maximum height. Then it looks like you used t/2 for the time in the Δx equation. That is, it looks like you used half the time to reach maximum height for the time in the Δx equation.

 

[emilytm]

OK, good.

OK, but I'm a little confused with your notation in your first post. You used the symbol t for the time to reach maximum height. Then it looks like you used t/2 for the time in the Δx equation. That is, it looks like you used half the time to reach maximum height for the time in the Δx equation.

Oh, I see what you mean. It should have been just t in the delta x equation and t/2 in the other equation, t representing total time.

 

[TSny]

Okay, I think I figured it out. I plugged in t=2Vo*sin theta/a into the equation delta y=Vyo*t+(1/2)at^2 and I got Vo=11.6 and t=2.37s.

I don't think it's possible to obtain Vo this way. Can you show more details of your calculation?

I do agree that the total flight time of the football is 2Vo*sin θ/|a|, where |a| is the magnitude of the acceleration.

If you want to use symbols such as θ, just click on the Σ symbol on the tool bar at the top of the window for entering your post.

This toolbar also has superscript and subscript icons, as well as other features.

 

[emilytm]

I forgot to square t. I think I am just going to get in person help on this. Thank you for your help!