# Homework Help: Archery Problem on Slope

1. Oct 8, 2007

### Umphreak89

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

An Archer standing on a 15 degree slope shoots an arrow 20 degrees above the horizontal. how far down the slope does the arrow hit if it is shot with a speed of 50m/s from 1.75m above the ground

2. Relevant equations

3. The attempt at a solution

y component = 50 * cos(20 + 15) - 9.8(t)
x component = 50 * sin (20 + 15)

not sure if the angles can be added..

if they can't then you'd have to do like 50cos(20) - 9.8t = -(1.75 + t*tan(15) ) with the arrow 's initial point being the origin. going from there, though, I'm not sure what to do

Last edited: Oct 9, 2007
2. Oct 9, 2007

### learningphysics

Do they give a picture? Is the archer standing on a 15 degree slope going upwards or a 15 degree slope going downwards?

3. Oct 9, 2007

### Umphreak89

slope is downwards

4. Oct 9, 2007

### learningphysics

ok. So what's the equation for horizontal displacement in terms of time? What's the equation for vertical displacement in terms of time?

5. Oct 28, 2007

### thirty-seven

I hope it is ok that I'm bumping this topic back up:uhh:

I'm having issues with the same problem:

I drew/scanned an image:

http://img150.imageshack.us/img150/6829/scannp3.jpg [Broken]

So what I have gotten is this:

d = starting d + initial v•t + 1/2•a•t(squared)

I made my starting distance 1.75m + the tan (15)•the horizontal distance.

my initial velocity is 50m/s so my vertical intitial velocity = 17.101m/s

my a = -9.81m/s(squared)

so...

0m = (1.75m + tan(15)•horizontal distance) + 17.101m/s•t + .5•-9.81m/s•t(squared)

i'm not sure what do from here...

Last edited by a moderator: May 3, 2017
6. Oct 28, 2007

### GTrax

Take this a little piece at a time.
Getting vertical and horizontal components of the arrow initial velocity is good - you need them.
But look at them carefully. I hope in your diagram, x directions are horizontal, and y are vertical.

You can just feel intuitively that an arrow shot at only 20 degrees above horizontal will be going faster horizontally than vertically. So re-visit your x and y components. Plug in the numbers and calculate them (even if temporarily). Make them make sense to you. Your arrow does not care what the slope of the land is, only where is horizontal!

Next is to know that however the arrow moved horizontally (it moved at a constant speed), it took exactly the same time as the vertical journey (up and down, while being accelerated downwards). Use this fact along with the equations of motion to get at the distance from the start point the arrow gets to. The 1.75m at the end is there just to mess with you a little.

Now you are into triangles. Pythagoras is OK. As you get into this, you could well notice that there is more than one way to do it.

Last edited: Oct 28, 2007
7. Oct 28, 2007

### turbo

Paraphrasing from GTrax's response: If you attack this as a problem of ballistics, you can simplify greatly. If you shoot a projectile horizontally, you need only compute its gravitational fall to figure out when and where it will impact the ground. If you shoot on a slope (either up-slope or down-slope) your point of impact will be higher than you expect because the horizontal component of the projectile's flight will be smaller than the total distance between the launcher and the point of impact.

8. Oct 28, 2007

### thirty-seven

my initial vertical velocity = sin(20) (times) 50m/s = 17.101m/s
my horizontal velocity = cos(20) (times) 50m/s = 46.98m/s

I know that my horizontal is
d=vt (or in terms of t: t = d/v)
so: t = d/46.98m/s

Getting my vertical equation in terms of t is where I am having the problem :-/
I pretty sure that I should use the equation:
Ending distance = starting distance + initial velocity (times) time + 1/2 (times) acceleration (times) time_squared

For that I am getting:
0 meters = (1.75m + tan(15) (times) horizontal distance) + 17.101m/s (times) time + 1/2 (times) -9.81 m/s_squared (times) time_squared ... when {x| x > 0}

I know I can set the times equal to each other but I don't know how :-/

I think...
t = .233726 ( (x + 62.1588)_to.1/2 + 7.4584) ...when {x| x > 0} t = positive
OR
t = - .233726 ( (x + 62.1588)_to.1/2 - 7.4584) {x| x > 0} t = negative

so...
.233726 ( (x + 62.1588)_to.1/2 + 7.4584) = d/46.98m/s

d = 287.5m

is this right? can someone look over my work please :)

9. Oct 28, 2007

### GTrax

OK - you have realized you had the sines and cosines swapped.
Now the vertical velocity is 50*sin(20) ..
and the horizontal velocity is 50*cos(20)

You got the vertical one calculated OK... $$\ u_h = 49.98$$
At this stage, you let go, and did not calculate the vertical initial velocity, which you will need. I am looking - but maybe I missed it.. ah ha, there its is 17.101 in the equation.
Anyways, getting to t = horiz_distance/49.98 is correct.

For the vertical bit, try not to get deflected by the starting distance. It has no place in the equation (yet) You are going to figure the (vertical) distance from the start point to where the arrow will end up, using the equation of motion..
$$s=ut+\frac{1}{2}g t^2$$

where $$u$$ is the initial vertical velocity.

Handle the signs carefully. Which direction you choose to be positive is arbitrary, the answer will come out right, so long as you stick to your choice. Suppose we say the distances downward from the launch point are positive. This makes the gravity downward acceleration positive also. The initial velocity (17.101 upwards) had better have a negative sign when you substitute that 'u'. Also, you decide a 'zero point' or 'origin'. It may be 1.75m up off the ground, so long as you remember that.

At this point, try and set these two conditions together. You know t in terms of horizontal distances and velocities, you can substitute it. Except for getting the initial velocity sign wrong, your vertical motion equation looks OK. The vertical distance from the launch point would be 1.75m plus what would have happened if the arrow had been launched from ground level

Using the 15 degree information, and that its tangent links the horizontal and vertical distances, is also useful here.

Last edited: Oct 28, 2007
10. Feb 19, 2008

### StephenDoty

I found how long it took the arrow to go up and come back down
But I had to have vf first

(17.101)^2 + (-19.6 * -1.75) = vf^2
326.74=vf^2
vf=-18.1
thus
-18.1=17.101- 9.8 t
-35.18= -9.8 t
t=3.59 s
then I times that t with the horizontal velocity of 46.98 m/s
and I got a range of 168.64m
but this is wrong

What did I do wrong?????

11. Oct 13, 2010

### tann_141

I have a really elegant solution for this problem. Your main issue is how to determine the vertical distance that the arrow will actually end up traveling. Well since you know the horizontal speed as being:
Vcos(a) (where a is the angle from the horizontal of the arrow being shot).

You can say that in 1s, the arrow will travel a horizontal distance of Vcos(a),

and the vertical distance will change by Vcos(a)tan(b) (where b is the angle of the slope from the horizontal)

What Vcos(a)tan(b) represents, is actually the "speed" at which the vertical component changes!

Now you have a model for the vertical distance that is related to time, simply by using d = V1t +1/2at^2, you can sub in the vertical "speed" in for d, remembering to take into account your initial height as well.

Vcos(a)tan(b) -h = -4.9t^2 + Vsin(a)t

h represents the initial height and -4.9 is 1/2 (g)... Vsin(a) represents the initial vertical height that the arrow has (opposite direction to g).

rearranging this equation, you get: 0 = -4.9t^2 + V(sin(a) + cos(a)tan(b))t + h

this is a quadratic that can be easily solved. Once you have t, you can multiply your horizontal speed Vcos(a) by t in order to get the total displacement in the x direction, then you simply divide by cos(b) in order to get the length of the hypotenuse, which is the distance from the archer to the point where the arrow has fallen.

** I worked it out with your values, and you should end up with 297.6 m **

Last edited: Oct 13, 2010
12. Oct 13, 2010

### zgozvrm

This problem can be solved by realizing that the path of the arrow describes a downward-facing parabola with a vertical line of symmetry and the hill describes a line.

Determine the equations for the parabola and the line and find the intersection.