(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!

See:

http://www.jiskha.com/display.cgi?id=1254088758

To solve this problem, we can break it down into two separate parts: finding the initial velocity and finding the launch angle for each archer.

Let's start with the archer at the bottom (archer A):

1. Finding the launch velocity (V1):
Since we know the distance between the archers (450m) and the angle of the shot (35 degrees), we can use the following kinematic equation for projectile motion:

d = V1 * cos(theta) * t

where d is the horizontal distance (450m), V1 is the launch velocity, theta is the launch angle (35 degrees), and t is the time of flight.

Now, we need to find the time of flight. Since the archer at the bottom is initially at the same height as the archer on top (150m), the vertical displacement (dy) will be zero. We can use the following kinematic equation to solve for t:

0 = V1 * sin(theta) * t - (1/2) * g * t^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Simplifying this equation, we get:

0 = V1 * sin(theta) * t - (4.9) * t^2

We can rearrange the equation to solve for t:

t = (V1 * sin(theta)) / 4.9

Substituting the value of t back into the equation for d, we have:

450m = V1 * cos(theta) * [(V1 * sin(theta)) / 4.9]

Solving this equation will give us the value of V1.

2. Finding the launch angle (theta) for archer A:
To find the launch angle for archer A, we can use the same equation we used to find t:

0 = V1 * sin(theta) * t - (1/2) * g * t^2

Since we already know the value of t, we can rearrange the equation to solve for theta:

theta = arcsin((0.5 * g * t^2) / V1)

Now that we have found the launch velocity (V1) and the launch angle (theta) for archer A, let's move on to the archer at the top (archer B).

3. Finding the launch velocity (V0):
Since the archer at the top is shooting downward, the angle of the shot will be different. We'll call this angle phi. To find the launch velocity for archer B, we can use the same approach as we did for archer A:

450m = V0 * cos(phi) * [(V0 * sin(phi)) / 4.9]

Solving this equation will give us the value of V0.

4. Finding the launch angle (phi) for archer B:
Using the same equation as before:

0 = V0 * sin(phi) * t + (1/2) * g * t^2

Rearranging the equation to solve for phi:

phi = arcsin((-0.5 * g * t^2) / V0)

Now that we have found the launch velocity (V0) and the launch angle (phi) for archer B, we have answered both parts of the question.

It's important to note that due to minor approximation errors and simplifications in the equations used, the values obtained may not be exact. Moreover, factors like air resistance have been ignored.