Posted by **Nikki** on Sunday, September 27, 2009 at 5:59pm.

(I'll try to describe this the best I can, it's a little hard without showing a picture)

This question deals with projectile motion:

An archer is standing 150 meters above another archer 450 meters away [so visually the 150m and 450m make the right angle]. The archer at the bottom is pointing his bow at 35 degrees.

How fast and at what angle should the archer at the top shoot the arrow in order to hit the archer at the bottom? (V0)

Likewise, how fast should the archer at the bottom shoot his arrow to hit the archer at the top? (V1)

----

I have no idea where to start on this. If it's any help, I tried to use Pythagorean Theorem and got 474.3m as the distance between the two archers. I also got 55 degrees for the angle at the top, but it may be off because the archer isn't standing at the very tip of the 150m wall. It's a weird question...but I know it's asking for velocity/angle.

Help please?

Thank you very much!

- Gorilla -
**MathMate**, Monday, September 28, 2009 at 8:59am
The variables in this question are:

1. Initial velocity of the arrow, v0, m/s

2. Angle with the horizontal, θ degrees

3. time to hit the target, t seconds.

There are two equations that govern these variables:

g is acceleration due to gravity = 9.8 m/s²

v0*cos(θ) = 450 m (horiz. distance) ....(1)

For the lower archer:

v0*sin(θ) + (1/2)(-g)t² = 150 m (vert. distance) .....(2)

For the upper archer:

v0*sin(θ) + (1/2)(-g)t² = -150 .....(3)

For the lower archer, the angle is determined, the remaining unknowns being v0 and t with two equations relating them, so there is a unique solution.

For the upper archer, the angle is undetermined. With three unknowns and two equations, there is an infinite sets of solutions. The archer will (arbitrarily) decide on the angle of attack, then the initial velocity can be determined.

In fact, the problem is very similar to the game called Gorilla included in the QBasic language with Windows 98.

I will proceed to solve the case of the lower archer.

From equation (1)

v0 = 450/(t*cos(θ))

substitute v0 in (2)

450/(t*cos(θ)) * sin(θ)*t + (1/2)(-g)t² = 150

Simplify to get:

(450*sin(theta))/cos(theta)-(g*t^2)/2=150

Solve for t:

t=10√3*√((1/g)(3tan(θ))

For θ=35°

t=5.8 seconds.

Substitute θ and t in (1) to get

v0 = 450/(t cos(θ))

=95 m/s

Please check my derivation logic and calculations.

## Answer This Question

## Related Questions

- Physics - (I'll try to describe this the best I can, it's a little hard without ...
- physics - An archer standing a horizontal distance d=50m away from a tree sees ...
- Physics - An archer stands a horizontal distance d= 55 m away from a tree sees ...
- physics - M1.) Nguyen jumps off a cliff with a height of 100 meters at a ...
- physics - M1.) Nguyen jumps off a cliff with a height of 100 meters at a ...
- physics - an archer at the edge of a cliff fires an arrow from a height of 14.6 ...
- physics - an archer at the edge of a cliff fires an arrow from a height of 14.6 ...
- Physics - At the other end of what would come to be known as the Roman Empire, ...
- Physics - At the other end of what would come to be known as the Roman Empire, ...
- Physics - At the other end of what would come to be known as the Roman Empire, ...

More Related Questions