An archer shoots an arrow at a 78.0 m distant target, the bull's-eye of which is at same height as the release height of the arrow.
(a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 34.0 m/s? (Although neglected here, the atmosphere provides significant lift to real arrows.)
Range = Vo^2*sin(2A)/g = 78 m.
34^2*sin(2A)/9.8 = 78
Solve for A.
449
20.695
To find the angle at which the arrow must be released to hit the bull's-eye, we can use the equations of motion for projectile motion.
Let's break down the problem into its components:
1. Horizontal motion: The arrow travels a distance of 78.0 m horizontally.
2. Vertical motion: The arrow is released and hits the ground at the same height (neglecting air resistance). We need to find the angle at which it must be released to hit the bull's-eye.
The equation for the horizontal motion is:
Horizontal distance (x) = Initial velocity (V0) * time (t) * cos(theta)
Here, we know the distance (x) and the initial velocity (V0). We need to find the time (t) and the angle (theta).
The equation for the vertical motion is:
Vertical distance (y) = Initial velocity (V0) * time (t) * sin(theta) - 0.5 * acceleration due to gravity (g) * (time (t))^2
In this case, the vertical distance (y) is zero because the arrow is released and hits the ground at the same height. The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Now, let's solve for time (t) in the horizontal equation:
t = Horizontal distance (x) / (Initial velocity (V0) * cos(theta))
We can substitute this value of time into the vertical equation:
0 = Initial velocity (V0) * (Horizontal distance (x) / (Initial velocity (V0) * cos(theta))) * sin(theta) - 0.5 * acceleration due to gravity (g) * ((Horizontal distance (x) / (Initial velocity (V0) * cos(theta))))^2
Now, we have an equation with the unknown angle (theta). We can solve it using numerical methods or approximation techniques.
When solving this equation, we find that the angle at which the arrow must be released to hit the bull's-eye is approximately 36.9 degrees.