(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)
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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!
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.
To solve this problem, you can break it down into two separate parts: finding the velocity and angle needed for the archer at the top to hit the archer at the bottom, and finding the velocity needed for the archer at the bottom to hit the archer at the top. Let's tackle each part step by step:
1. Finding the velocity and angle for the archer at the top (V0) to hit the archer at the bottom:
- First, it's important to note that both archers are subject to the same gravitational acceleration, which we'll assume is approximately 9.8 m/s².
- To calculate the initial vertical velocity component of the projectile launched by the archer at the top, you can use the trigonometric relationship of the projectile motion. Since the archer is aiming at a 35-degree angle relative to the horizontal, you can calculate the vertical component using the equation V0y = V0 * sin(θ), where V0y is the vertical component of V0 and θ is 35 degrees.
- Next, you can calculate the horizontal component of the initial velocity using the equation V0x = V0 * cos(θ), where V0x is the horizontal component of V0.
- Since the initial vertical velocity component and the time of flight for the projectile are the same for both archers (as they are at the same vertical height), you can use these values to find the total flight time (t) of the projectile based on the vertical displacement using the equation y = V0y * t + (1/2) * g * t², where y is the vertical displacement (150 m) and g is the acceleration due to gravity (approximately -9.8 m/s² since it acts downward).
- With the total flight time, you can then calculate the horizontal displacement (range) of the projectile using the equation x = V0x * t, where x is the horizontal displacement (450 m).
- Now you have two equations describing the projectile motion: "y = V0y * t + (1/2) * g * t²" and "x = V0x * t". You can solve these two equations simultaneously to find V0 (the initial velocity magnitude of the projectile) and t (the total flight time).
- Once you have V0, you can find the launch angle (θ) by using the equation θ = arctan(V0y / V0x), where arctan is the inverse tangent function.
2. Finding the velocity for the archer at the bottom (V1) to hit the archer at the top:
- This can be approached similarly to the first part. First, let's assume the archer at the bottom is aiming at an angle of θ1.
- Using the same equations as before, you can find the vertical and horizontal components of V1: V1y = V1 * sin(θ1) and V1x = V1 * cos(θ1).
- Now, in order to hit the archer at the top, the archer at the bottom must launch the projectile simultaneously with the archer at the top, and the time of flight for the projectile should be the same as in the previous part.
- Using the equation y = V1y * t + (1/2) * g * t² (since the vertical displacement is now -150 m), you can solve for V1.
- Once you have V1, you can find the launch angle (θ1) by using the equation θ1 = arctan(V1y / V1x).
By following these steps, you should be able to find the velocities (V0 and V1) and angles (θ and θ1) needed for both archers to hit their targets.