Projectile motion of a football punt

[bob1182006]

Homework Statement

A football player punts the football so that it will have a "hang time" (time of flight) of 4.50s and land 50 yd (=45.7m) away.
If the ball leaves the player's foot 5.0 ft (=1.52m) above the ground, what is it's initial velocity (magnitude and direction)?

Homework Equations

x=v_{0x}t

v_x=v_{0x}
v_y=gt

v_0=v

\theta_0=\theta+\pi

v=\sqrt{v_x^2+v_y^2}

\theta=arctan\frac{v_y}{v_x}

The Attempt at a Solution

plugging into solve for x and y component of velocity I get:

v_x=v_{0x}=\frac{x}{t}=\frac{45.7m}{4.50s}=10.16 m/s
v_y=gt=-9.81m/s^2 * 4.50s=-44.15m/s
v=\sqrt{v_x^2+v_y^2}=\sqrt{(10.16)^2+(-44.15)^2}=\sqrt{2052}=45m/s

\theta=arctan\frac{-44.15}{10.16}=-76.5 degrees

initial velocity =45 m/s
angle is 76.5 degree's.

but my book says this is wrong >< degrees should be ~65, velocity about 35 m/s

Am I using the wrong equations?

 

[learningphysics]

The way you calculated vy doesn't make sense...

 

[Mindscrape]

I would look at your answer for v_y again. Are you sure this makes sense, and is what you want?

 

[bob1182006]

well the acceleration is 9.81 m/s^2 downward. time is 4.5s I want to end up with m/s which I do get.

Why does it not make sense?
it says that the ball will be going in a downward path since the v_y is negative.

What way should I have calculated v_y?

hm..well maybe I have the signs wrong? v_y=-gt? but that still gives the same answer.

 

[learningphysics]

well the acceleration is 9.81 m/s^2 downward. time is 4.5s I want to end up with m/s which I do get.

Why does it not make sense?
it says that the ball will be going in a downward path since the v_y is negative.

What way should I have calculated v_y?

hm..well maybe I have the signs wrong? v_y=-gt? but that still gives the same answer.

Are you trying to get the initial v_y or the final v_y ? also v = v0-gt. you're assuming v0 = 0

 

[Mindscrape]

Where is the ball kicked from? How much of that 4.5s time interval that you are using applies to the downward velocity you are talking about?

 

[bob1182006]

final v_y since from that I can get the final velocity and direction, and from that find the initial.

The problem doesn't state if the player drops the ball and then punts or punts it while it's on the ground having v0=0. so I'm assuming v0=0.

the ball is kicked from x=0, and that velocity would apply from when the ball is at it's peak until it hits the ground. the velocity is increasing from t=0 to t=time to reach the peak.

 

[learningphysics]

final v_y since from that I can get the final velocity and direction, and from that find the initial.

The problem doesn't state if the player drops the ball and then punts or punts it while it's on the ground having v0=0. so I'm assuming v0=0.

That doesn't work, since the ball needs an initial vertical velocity (right when it is kicked)... otherwise there would be no flight... you can't assume v0=0.

 

[bob1182006]

but wouldn't the initial velocity be equal in magnitude to the final velocity except the angle of the final is (initial angle + 180*)?
so finding the final velocity will allow me to find the initial velocity?

Or how would you do this? how should I start? finding the initial x-y components?

so initial x is the same, but initial y would be v0 and final would be v0-44.15m/s

and I don't know how you could find v0 there.

Also is that "the ball leaves the player's foot 5 ft above the ground" useless information?

 

[learningphysics]

but wouldn't the initial velocity be equal in magnitude to the final velocity except the angle of the final is (initial angle + 180*)?
so finding the final velocity will allow me to find the initial velocity?

Or how would you do this? how should I start? finding the initial x-y components?

so initial x is the same, but initial y would be v0 and final would be v0-44.15m/s

and I don't know how you could find v0 there.

Also is that "the ball leaves the player's foot 5 ft above the ground" useless information?

No, you need to use that. Just use the equation for vertical displacement in terms of time. vertical displacement is -5ft = -1.52m.

 

[bob1182006]

>< wow sorry didn't see that...

so -1.52m=v_{0y}(4.50s)-4.91\frac{m}{s^2}(4.50s)^2
which gives v_0y=21.8m/s and combining with v_0x to find v_0 gives 24.1m/s
and angle of 64.9 degrees N of E.

thanks all now I am going to go try another problem like this.

 

[learningphysics]

>< wow sorry didn't see that...

so -1.52m=v_{0y}(4.50s)-4.91\frac{m}{s^2}(4.50s)^2
which gives v_0y=21.8m/s and combining with v_0x to find v_0 gives 24.1m/s
and angle of 64.9 degrees N of E.

thanks all now I am going to go try another problem like this.

no prob.