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[BlackSalad]

I need help =(

A quarterback throwns a football straight toward a receiver with an initial speed of 21.0 m/s, at an angle of 25.5° above the horizontal. At that instant, the receiver is 17.5 m from the quarterback.

With what constant speed should the receiver run to catch the football?

HINT he will have to run in the direction the ball was thrown

now I figure I need to find the delta x first, and I've tried it in two different ways and came up with ~19.5996 m one way and 30.160 m for the other but I don't think either are right. help me GAF!

 

[Dilbert]

1) Calculate the time of flight of the football. You will use your standard 1-D kinematic equations: the initial velocity is equal to 21.0 sin 25.5° (the vertical component of the velocity vector), the initial and final displacement is 0 (level playing field), and acceleration equals -9.80 m/s² (assuming sea level).

2) Calculate the horizontal distance the football travels. Again, use 1-D kinematics: the initial displacement in the x-direction can be assumed to be 0, the time of flight is the answer you found in part 1, and the horizontal component of velocity is 21.0 cos 25.5°.

3) Calculate the horizontal distance the receiver has to travel. Subtract his starting position (17.5 m) from the answer to part 2.

4) Calculate the speed of the receiver. Divide the answer to part 3 by the answer to part 1 and your problem should be solved.

One more thing -- if your answer is in meters, you have something screwed up in your dimensional analysis. It should be m/s for this problem.

 

[Drozmight]

You need to figure out how far the ball will go first and subtract the 17.5 m from that to get the total distance the reciever needs to travel. Probably need to find out how long the ball will be airborne and then find out when the Sf (final position) for both objects are equal to one another.

The equation for the ball escapes me the other side of the quation for the quater back would just be deltaT * X.

Edit: Or what jinx said.

 

[BlackSalad]

ack, frustrating, thats what I thought I did =(

Here is what i have so far:

t = 1.53882
xfinal = 30.160 m
velocity of the reciever = 8.22727

I'm stuck >_<

thanks for the quick replies, btw

 

[Drozmight]

for the distance of the ball I got (21^2)*sin(2*25.5) / 9.8 = 34.97m
for the time I got 2(21)(sin 25.5) / 9.8 = 1.85s

34.97 - 17.5 = 1.85 * Vs

Vs = 9.4432 m/s

Don't know if I did that right, I haven't done trajectory in a long ass time.

 

[Dilbert]

1) y = y0 + v0t + ½at²

Since y = y0 = 0, you get (after factoring):

0 = t(v0 + ½at)

t = 0 is the trivial solution (it's at the point you throw the ball), so solve for the second term:

v0 = -½at
t = -2v0/a
t = -2(21 sin 25.5°)/(-9.80)

t = 1.85 s that the ball is in the air

2) For 1-D kinematics with no acceleration:

x = x0 + v0t

Assuming that x0 = 0:

x = v0t
x = (21 cos 25.5°)(1.85)

x = 35.1 m that the ball travels horizontally

3) How far does the receiver have to run? Well, duh...

35.1 m (ball) - 17.5 m (receiver headstart) = 17.6 m that the receiver has to run while the ball is in flight

4) More 1-D kinematics with no acceleration...you've seen this formula before:

x = v0t
v0 = x/t
v0 = (17.6 m)/(1.85 s)

v0 = 9.51 m/s

Voila, presto, etc.

 

[BlackSalad]

my calculator was on radians

doh!

 

[Drozmight]

haha.

That dude's gotta book it to catch that ball, damn.

 

[BlackSalad]

I don't think it'll catch it

thanks for the help guys!

 

[alejob]

One last thing, if the QB is Donovan McNabb, throw your calculations away cos its gonna get picked of.

 

[WHOAguitarninja]

JINX!!!! YOU'RE MY LAST HOPE!!!(I have a question also)

Okay, here goes.

I have a solid sphere of radius a. Inside is a sphere carved out of radius a/2.
Looks like this...

The solid portion is filled with a charge of uniform density p(thats rho, but i guess it doesn't really matter).

I need to find the strength of the electric field at points B and C(assuming C is inside the empty space...kindof messed up in my drawing).

I tried using an integral which started complicated but simplified fairly nicely but i had infinites popping up everywhere when i tried to solve it.

I'm 99% sure you need to use Gauss's law, but I don't know how.

 

[Dilbert]

JINX!!!! YOU'RE MY LAST HOPE!!!(I have a question also)

Okay, here goes.

I have a solid sphere of radius a. Inside is a sphere carved out of radius a/2.
Looks like this...

The solid portion is filled with a charge of uniform density p(thats rho, but i guess it doesn't really matter).

I need to find the strength of the electric field at points B and C(assuming C is inside the empty space...kindof messed up in my drawing).

I tried using an integral which started complicated but simplified fairly nicely but i had infinites popping up everywhere when i tried to solve it.

I'm 99% sure you need to use Gauss's law, but I don't know how.

Actually, it's a fairly simple problem...if you know the trick. Remember that from the perspective of any point in space outside of a spherical volume V with uniform charge density p, it "looks like" the field you'd get from a point charge with the equivalent total charge pV located at the center of the sphere. With that in mind...here's the trick.

Mathematically speaking, the area of empty space is equivalent to the superposition of an area of NEGATIVE charge density rho over the area of positive charge density rho. (When you sum them, there is zero charge in any area inside the spherical cavity...which matches the physical reality.) From any point outside the large semi-hollow sphere, the electric field is identical to the vector sum of two point charges, one centered at (0,0) with total charge (4/3)(rho)(pi)a³, and one centered at (0, a/2) with total charge -(4/3)(rho)(pi)(a/2)³.

Hopefully that will give you a kick in the right direction. Otherwise, I'm useless with EM -- I hated that subject the most when I went through physics.

 

[WHOAguitarninja]

AH!!! I WANT TO HAVE YOUR BABIES THANK YOU YOU'RE AWESOME!!!!

The superposition of the empty space as negative to cancel it out is the part I was missing. Thank you SOOO much. You have no idea how much that just cleared up for me.

 

[nitewulf]

i think that's pretty much it. with spherical conductors, gauss' law should simplify to E=Q/(4*pi*R^2).
you are dealing with two points, one inside the big sphere, one outside.
the electric field at the point inside the big sphere would only be affected by the hollow sphere.
and the electric field at the point outside will be affected by charges from both spheres, as it lies outside both spheres.

 

[WHOAguitarninja]

One last question...How is it that the point at a is only affected by the hollow sphere when there is a lot of positive charged mass to the side of it? Does that stuff just not do anything?


EDIT- One more question too....I'm using super position to find B, by adding the two different q/r^2 values. But I'm feeling I'm doing something wrong cuz I'm not using Gauss's law.

 

[nitewulf]

One last question...How is it that the point at a is only affected by the hollow sphere when there is a lot of positive charged mass to the side of it? Does that stuff just not do anything?

Is the point A inside the solid sphere, lying at a greater radius from the hollow sphere?
As per Gauss' Law, you know that conductors have an Electric field of zero within. But it does lie outside the hollow sphere, so it is affected by it.

EDIT- One more question too....I'm using super position to find B, by adding the two different q/r^2 values. But I'm feeling I'm doing something wrong cuz I'm not using Gauss's law.

Are you using the E equation I reffered to? If so, then it is Gauss' law, for a sphere. I did forget to include "epsilon zero" or the premittivity of free space, it should be E=Q/(4*pi*e0*R^2)

 

[WHOAguitarninja]

Ohhhh yea.


Thanks so much guys, you saved my week, seriously.