Physics (repost)
posted by Nikki .
(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
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.
Respond to this Question
Similar Questions

Physics
(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 … 
Physics
At the other end of what would come to be known as the Roman Empire, Parthian (Persian) archers were able to discourage pursuers by firing backwards from the back of a galloping horsethe famous "Parthian shot." a. If the speed of … 
Physics
At the other end of what would come to be known as the Roman Empire, Parthian (Persian) archers were able to discourage pursuers by firing backwards from the back of a galloping horsethe famous "Parthian shot." a. If the speed of … 
physics
an archer at the edge of a cliff fires an arrow from a height of 14.6 m above the ground at the bottom of the cliff. the arrow just clears a rock on a pillar that is 21.7 m above the gound which is 45.2 m horizontally away from the … 
physics
an archer at the edge of a cliff fires an arrow from a height of 14.6 m above the ground at the bottom of the cliff. the arrow just clears a rock on a pillar that is 21.7 m above the gound which is 45.2 m horizontally away from the … 
Physics
An archer stands a horizontal distance d= 55 m away from a tree sees an apple hanging from the tree at h= 12 m above the ground. The archer chooses an arrow and prepares to shoot. The arrow is initially 1.5 m above the ground. Just … 
physics
An archer standing a horizontal distance d=50m away from a tree sees an apple hanging from the tree at h=12m above the ground. The archer chooses an arrow and prepares to shoot. The arrow is initially 1.5m above the ground. Just as … 
physics
M1.) Nguyen jumps off a cliff with a height of 100 meters at a velocity of 3.5 meters per second to a save a cat. How long will it take her to reach the surface of the ocean? 
physics
M1.) Nguyen jumps off a cliff with a height of 100 meters at a velocity of 3.5 meters per second to a save a cat. How long will it take her to reach the surface of the ocean? 
Physics
An archer lying on the ground shoots an arrow at a target that is 400 meters away, and 40 meters high. The initial velocity of the arrow is 120 m/s At what angle does he need to point his bow in order to hit the target?