To solve this problem, we can use the kinematic equations of motion for vertical motion:
1. The equation for the final velocity, vf, in terms of initial velocity, vi, acceleration due to gravity, g, and time, t, is:
vf = vi - gt
2. The equation for the displacement, d, in terms of initial velocity, vi, acceleration due to gravity, g, and time, t, is:
d = vi * t - 1/2 * g * t^2
For the first arrow:
- Initial velocity (vi1) = 37.5 m/s
- Time taken to reach maximum height (th1) = ?
- Acceleration due to gravity (g) = 9.8 m/s^2
Using equation 1, we can say that at maximum height, the velocity becomes zero:
0 = vi1 - g * th1
Solving for th1, we have:
th1 = vi1 / g
Now we'll solve for the second arrow. Let's assume the initial velocity of the second arrow is vi2, and it is fired 1.66 s after the first arrow.
For the second arrow:
- Initial velocity (vi2) = ?
- Time taken to reach maximum height (th2) = th1 - 1.66 s
Using equation 1 again, we have:
0 = vi2 - g * th2
Substituting the value of th2, we get:
0 = vi2 - g * (th1 - 1.66)
Solving for vi2, we have:
vi2 = g * (th1 - 1.66)
Now, substituting the value of th1 = vi1 / g, we get:
vi2 = g * ((vi1 / g) - 1.66)
Simplifying:
vi2 = vi1 - 1.66 * g
Plugging in the values of vi1 = 37.5 m/s and g = 9.8 m/s^2:
vi2 = 37.5 - 1.66 * 9.8
Calculating vi2:
vi2 = 37.5 - 16.268
vi2 ≈ 21.232
Therefore, the initial speed of the second arrow is approximately 21.232 m/s.