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.