# Homework Help: Projectile Motion (shooting an arrow)

1. Oct 29, 2007

### KatieLynn

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

An arrow is shot at 42 degrees from the horizontal to strike a target that is 53 meters away and at the same elevation as the archer. What is the initial velocity?

2. Relevant equations
I'm not sure which equation(s) is/are needed so I'll list all the ones I can...
(Vix)=(CosX)
x=(Vix)(t)
(Vxf)=(Vxi)
(-Vi)(SinX) = (Vi)(SinX) + (g)(t)
(-2Vfy)=(g)(t)
(Vfy)^2=(Viy)^2 + 2(g)(y)

g=-9.81 m/s

3. The attempt at a solution
I don't know how to solve it. All the information you are given is theta = 42 and the X distance is 53 m, I don't see how you can use any of those equations to solve for anything with only those two pieces of information.

2. Oct 29, 2007

### Kurdt

Staff Emeritus
The only equations you should need are the kinematic equations which are given here:

The trick with these problems is to treat the two components of motion separately. You know the x component of velocity will be $u_x=u\cos \theta$ (I've used u for initial velocity because v is usually used for final velocity) and I think you've identified that the distance will be given by $s=u_xt$ (where I've used s in place of your x). All you need to do is find an expression for t which is where the vertical motion comes into it.

What can you write for the vertical motion?

3. Oct 30, 2007

### KatieLynn

Vxi=VicosX

X= Vix (t)

so when you plug in numbers and solve for t you get..

t= (Vi*Cos42)/(53)

then if you substitute that equation for t into

-2Vi(sinX)=gt

you get

-2Vi(sin42)=(-9.81)(Vicos42/53)

which equals 9.7

I'm a little worried thought because the answer my teacher gave me was 8.85

is that close enough, maybe a little off because of rounding, or is it just wrong?

4. Oct 30, 2007

### pinkyjoshi65

The equation below is right, so if you want to solve for t, it will be, t=X/Vix not Vix/X
X= Vix (t)

So, there is a mistake in this equation below..
-2Vi(sin42)=(-9.81)(Vicos42/53)

5. Oct 30, 2007

### Kurdt

Staff Emeritus
As pinkyjoshi points out you have calculated t incorrectly. Even more perplexing is the fact that I get a different answer to your teacher.

6. Oct 30, 2007

### KatieLynn

Thats not good...not good at all.

7. Oct 30, 2007

### Kurdt

Staff Emeritus
Ok so now you have $t = \frac{s}{ucos \theta}$ and you've subbed it into:

$$\frac{2usin \theta}{g} = t$$

If you rearrange in terms of u and plug the numbers in you do get a different value from the one your teacher says is correct. I suspect they've made an error or given the answer to something else etc.