# Projectile Motion and initial speed

1. Sep 21, 2007

### devilsangels287

1. The problem statement, all variables and given/known data

Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.20m/s . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 48.8m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 4.50 secs after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.

Q: With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?

Q:Where is Henrietta when she catches the bagels?

2. Relevant equations

I placed the origin to be at the top of the picture. Where 48.8 would become 0. (I hope you understand what I mean)

So I know:

Vox= 4.20 m/s Voy= ?
Xo= 0m Yo = 0m
X = ? Y= -48.8m
g= 9.8 m/s^2
t= 4.5 secs t= 4.5 secs

I think that the x and y time are the same.

3. The attempt at a solution

So what I did was first find the x, so

X= Vox*t
X= 4.2*4.5
X= 18.9m

So that was suppose to answer part B question, but its wrong so I don't how else to do it.

And then for the part 1 question I did this

Y=Yo + Voy*t - (1/2)(g)(t^2)
-48.8= 0 + 4.5(Voy) - (1/2)(9.8)(4.6^2)
-48.8= 4.5(Voy) - 99.225

50.425 = 4.5Voy
Voy= 11.206 m/s

So I also thought that was correct, but it wasn't and don't any other way to do this. Please help me! Thanks.

2. Sep 21, 2007

### learningphysics

The part you're getting wrong is, he throws the lunch after 4.5.... she doesn't catch them after 4.5.

set t = 0, to be when he throws the lunch... where is Henrietta at this time?

3. Sep 22, 2007

### devilsangels287

If I were to plug t = 0, wouldn't it make the answer equal to 0 then?? Am I using the right equations for this problem? Because these are the only equations that my teacher gave me.

X= 4.2(0)= 0m

Y=Yo + Voy*t - (1/2)(g)(t^2)
-48.8= 0 + 0(Voy) - (1/2)(9.8)(0^2)
-48.8= ?

I don't know if I am using the right equations?

4. Sep 22, 2007

### tabchouri

answear the issues in the problem bit by bit.

Setting t=0, is just a shift for the sake of computational simplicity, just as when you set Y=0 instead of 48.8. Then you can convert time to relative when she started running.

now, answear this : when the husband throws the lunch, where is his whife ? (coordinate Xh0,Yh0)

5. Sep 22, 2007

### devilsangels287

I don't know?? 0m?? I mean if I using the correct equation wouldn't it be zero because we are assuming that t= 0

X= 0+(4.2)(0)
X= 0??? Is this correct?

6. Sep 22, 2007

### drpizza

There are two components of the velocity: the vertical component and the horizontal component. Work with these independently. Use the vertical component to calculate the time until "just before they hit the ground." It doesn't matter if she's running 2 m/s or 10 m/s - either way, it's going to take just as long for the bagels to reach the ground.

7. Sep 22, 2007

### tabchouri

no, the husband throws the meal at $$t=4.5$$
so the wife is at
$$X_{w0} = 0 + 4.2 * t$$

and in all the experiment, the height of the wife is at the ground wich is normally 0. but you shifted it so that the window's (in the appartment) heigh = 0.
so the heigh of the wife is
$$Y_{w0} = -48.8 m$$

Further, the coordinates of the meal when it is lauched are ($$X_{m0} = 0$$, $$Y_{m0}=0$$), right ?

Now that is the conditions when the meal is thrown, at t=4.5.

We can reconsider another shifted time, $$t' = 0$$ when $$t=4.5$$.

You can solve the equations from there, taking into accout $$t'$$ and not $$t$$.

-----------------------------------------------------
Correct me if I am wrong.

8. Sep 22, 2007

### learningphysics

Henrietta is at 18.9m when he throws the bagels... So Xo = 18.9m.

So we are setting the start point of the problem at the moment he throws the bagels... as tabchouri explained... at this point in time... For the bagels... Xo=0. Yo=0. For Henrietta Xo = 18.9. Yo=-48.8

9. Sep 23, 2007

### devilsangels287

So when t= 0 secs, Henrietta is 18.9m away from the house. So we have to find what time she catches the bagels to find the additional distance that she is away from the house?

10. Sep 23, 2007

### learningphysics

yes... and to get the velocity with which the bagels are thrown.

11. Sep 23, 2007

### devilsangels287

I find the time of the bagels to drop, which was 3.16secs.
I used the equation t=sqrt((2Y_0)/g)) = sqrt((2*48.8)/9.8) = 3.16 secs

Then since we know the time that it took Henrietta to walk down 4.5 secs + 3.16 secs(the time it took the bagels to drop) now equals to 7.66 secs

X= X_o+V_ox*t
X= 18.9 + 4.2*7.66
X= 51.072m

Is that the distance from the house where Henrietta catches the bagels?
Is that the right way to do it?

If so, how would I do find the velocity?

12. Sep 23, 2007

### learningphysics

Okay, so you're using the t = 0 point to be when she's right at the building... ie x = 0.

If you use the t = 0 point to be when she's right at the apt. building, then Xo = 0m

X = 0 + 4.2*7.66 = 32.172m

If you use the t = 0 point to be when the bagels get thrown, then Xo = 18.9m

X = 18.9 + 4.2*3.16 = 32.172m

Distance divided by time... 32.172m/3.16s

13. Sep 23, 2007

### devilsangels287

thanks so much for your help!!