# Projectile Question

1. Oct 27, 2008

### harrycc

1. The problem statement, all variables and given/known data
An Archer is planning show an arrow at a target 125 meters aways in which the center is 2 meters above the ground. The arrow leaves his bow with a speed of 50m/s at an angle of 15 degrees and a height of 1 meter off the ground.

a)How close does he come to hitting the center of the target. i.e. at what height does it impact the target?
b) how long was the arrow in the air?
c) With what velocity (magnitude and direction) does the arrow hit the target.

2. Relevant equations
V = Vi + at
change of x = Vit + 1/2(at^2)
V^2 = Vi^2 + 2a(change of x)

with component factor

3. The attempt at a solution

I found the correct answer of b which is 2.65 seconds and I use it for the second equation to solve for Y final. I got 1.526m for Y final but my instructor said the answer is not correct. Where did I made a mistake and what is the correct way to solve for this problem.

2. Oct 28, 2008

### phyzmatix

It will be easier to comment if you show us what you did. Did you resolve your projectile motion into its vertical and horizontal components?

3. Oct 28, 2008

### harrycc

this is what I did.

Solve for T

Change of X = V initial (x component) * t
125 = (cos15)(50)t
t = 2.6s

That is the first part and answer for (B), then I tried to solve for A through through the following steps:

Change of Y = V initial (Y component) t + 1/2 a(component y) * t^2
Y - 1m = (sin15) * (50) * (2.6) + (1/2)(-9.8)(2.6)^2
Yfinal = 1.526m

Please explain how to get the right answer. Thank You

4. Oct 28, 2008

### DarkSamurai

The best way to setup these problems is first create a table

X_initial
X_final
Vx_initial
Vx_final
Ax
time

Y_initial
Y_final
Vy_initial
Vy_final
Ay
time

The two things that we know is that acceleration is going to be 0 in the x-direction in these types of problems (always). And that the gravity in the y direction will be the earth's gravity in this case, but could be any gravity.

The third thing is to note that time will always be the same in the x and y direction. So this can be a link =).

Good luck.

5. Oct 28, 2008

### harrycc

Yes, I am very sure that is what I did, but however my answer for part A is not correct and I dont know why. Please help

6. Oct 28, 2008

### alphysicist

Hi harrycc,

I think its confusing, but I believe in part a they are wanting you to find out how far the arrow is from the center of the target. You found that at the end the arrow is 1.526m from the ground. But since the problem first asks: how far is the arrow from the center of the target, and the center is 2m off the ground, what would that be?

(It's confusing though because right after that it mentions the height, which would seem to be the distance from the ground.)

(Also, if significant figures are important, you might want to keep more digits while you are doing your work. Your result 1.526 is over 8% away from the true value. So you said the time was 2.6s, but I would definitely keep more digits than that.)