# Arrow Being Launched Distance

1. Oct 7, 2008

### patv

[SOLVED] Arrow Being Launched Distance

1. The problem statement, all variables and given/known data
An Arrow is launched with an initial velocity of 12.0m/s at an angle of 22.0 degrees with the horizontal, and is released at a height of 2.5m above the ground. How far away does it land on the ground?

2. Relevant equations
d=vit + $$\frac{1}{2}$$at2
a =$$\frac{vf - vi} {t}$$ (I couldn't get this to display properly, but where it has sub in square brackets, it should be subscript.)

3. The attempt at a solution
$$\sin{22}$$ =$$\frac{vy}{12}$$
4.5m/s=vy
$$\cos{22}$$ = [ vx ] over 12 (I got an error the first time I tried this in latex, sorry)
vx = 11.14 m/s

After getting the velocity in the x and y direction, I attempted to find the height above the original shooting height the arrow would reach. I used a formula I cannot recall at the moment, and got 1m, but was informed that this was not correct.
Next, I will attempt to find the time the arrow will be in the air.
I tried to sub vt into the place of d in the equation below, and this gave me a time of 2.177 seconds. The substitute teacher said this was incorrect, but did not give a reason.

d=vit + $$\frac{1}{2}$$at2

I know I'm supposed to find time first, but I really am not sure of how to get this. Can anyone provide assistance, or insight as to which formula to use, or where I am going wrong?
Thanks
Below is a diagram of how I think it would be drawn.
http://img60.imageshack.us/img60/9348/fw3tixcm6.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

Last edited by a moderator: May 3, 2017
2. Oct 8, 2008

### Rake-MC

Okay, so a good start by splitting the initial velocity into two components (x,y).

The next step you'll want to do is calculate how long it's in the air for. So define the y point that it's launched from as zero.
You can use the formula: $$s = ut + \frac{1}{2}at^2$$ (where s = displacement, a = acceleration, u = initial velocity and t = time) to calculate the time it's in the air.
Such that s = -2.5, u = 4.5, a = -9.81 and you're solving for t.

We also know that there is no horizontal acceleration, therefore velocity is constant.
You have initial velocity, time, acceleration and you're solving for distance.

Once again you can use $$s = ut + \frac{1}{2}at^2$$ however a = 0 therefore it becomes $$s = ut$$

Good luck.

3. Oct 8, 2008

### Kurdt

Staff Emeritus
Finding the time with the distance equation like you tried is the way to go. What you neglected to do however is take into account the arrows initial launch height. The more general form of that distance equation is:

$$s = s_{0} +ut+\frac{1}{2} at^2$$

(You can click on my latex image to see how to render subscripts and superscripts correctly)

4. Oct 8, 2008

### patv

Thanks guys!
I ended up getting the correct answer, with your help. I can't recall the answer, I don't have it here, but it was correct.
Thanks!