# Archer on a cliff

1. Sep 19, 2006

### dboy83

An archer stands on a cliff that is 100m high, shooting at a mark on the ground below 430m away from the bottom of the (vertical) cliff. The archer is capable of shooting an arrow with an initial speed of 35 m/s at any angle you choose.

A) If the archer shoots horizontally ( at an angle of zero degrees with respect to the horizontal), will the arrow reach the mark? why? If not, will shooting at any angle above the horizontal help? Why or why not?

B) What initial speed of the arrow is needed to be in order for the arrow to reach the target mark ( still with an angle of zero degrees)?:rofl:

2. Sep 20, 2006

### Staff: Mentor

You need to show your work so far. What equations of motion apply to this problem?

3. Sep 20, 2006

### Delzac

Just some hints, use your kinematics equations, the 4 equations. And the time for vertical motion and horizontal motion is the same.

Hope it helps

4. Sep 20, 2006

### dboy83

lost...

I don't know which one to use. Where do i start? Please somebody help.

5. Sep 20, 2006

### dboy83

what formula

does anybody know what formula I am supposed to use?

6. Sep 20, 2006

### Staff: Mentor

What is the equation that relates changes in velocity to the initial velocity and the acceleration? What is the slightly longer equation that relates changes in position to initial position, initial velocity, and acceleration?

7. Sep 20, 2006

The velocity in the x direction is constant and it equals the initial velocity v0 = 35 m/s. The y direction velocity equals gt. Hence, the path in the x direction equals v0*t = 35*t (1), and in the y direction 0.5*g*t^2. Now if the arrow falls to the ground, the path that it has to cross in the y direction equals 100 m, so 0.5*g*t^2=100. Solving this equation gives us the time t of the flight of the arrow. Putting that time into equation (1) gives us the x displacement of the arrow, so we can tell if it has his the target or not.

8. Sep 20, 2006

### dboy83

hmm

You guys make it look so easy... Ok, so 0.5*g*t^2=100

I plug in -9.80 for g and multiply that by 0.5, which gives me -4.9.
so I could rewrite the equation as -4.9*t^2=100 If I try to solve for t I get the square root of a negative number which is not possible. ?????

9. Sep 20, 2006

### dboy83

I'm thinking

is 0.5*g*t^2=100 supposed to be 0.5*g*t^2=-100 ? ( the only difference is the negative sign in the 100.

10. Sep 21, 2006