an archer shoots an arrow at 83.0 m/s at a 62.0 degrees angle to the horizon. How much time passes before the arrow reaches the ground?

To find the time it takes for the arrow to hit the ground, we need to break down the arrow's initial velocity into its horizontal and vertical components, and then use the kinematic equations for projectile motion. Here’s a step-by-step solution for the given problem:

1. **Initial Velocity Components:**
- The initial velocity \( v_0 = 83.0 \) m/s.
- The launch angle \( \theta = 62.0 \) degrees.

The horizontal component of the initial velocity:
\[
v_{0x} = v_0 \cos(\theta)
\]
The vertical component of the initial velocity:
\[
v_{0y} = v_0 \sin(\theta)
\]

Using the provided values:
\[
v_{0x} = 83.0 \cdot \cos(62.0^\circ) \approx 39.0 \, \text{m/s}
\]
\[
v_{0y} = 83.0 \cdot \sin(62.0^\circ) \approx 73.2 \, \text{m/s}
\]

2. **Vertical Motion Analysis:**
We will use the kinematic equation that incorporates the initial vertical velocity, acceleration due to gravity, and the displacement to determine the total time of flight. The key is to set the displacement \( y \) equal to zero when the arrow returns to its launch height (i.e., when it hits the ground again).

The relevant kinematic equation for vertical displacement \( y \) as a function of time \( t \) is:
\[
y = v_{0y} t - \frac{1}{2} g t^2
\]
Here, \( y = 0 \) (since the initial and final heights are the same), \( g \approx 9.8 \, \text{m/s}^2 \), and \( v_{0y} \approx 73.2 \, \text{m/s} \).

Setting \( y = 0 \), we get:
\[
0 = 73.2 t - \frac{1}{2} \cdot 9.8 \cdot t^2
\]

To solve this quadratic equation, factor out \( t \):
\[
0 = t \left( 73.2 - 4.9 t \right)
\]

This provides two solutions:
\[
t = 0 \quad \text{or} \quad 73.2 - 4.9 t = 0
\]

Solving for \( t \) when \( 73.2 - 4.9 t = 0 \):
\[
73.2 = 4.9 t
\]
\[
t = \frac{73.2}{4.9}
\]
\[
t \approx 14.94 \, \text{s}
\]

Thus, the time it takes for the arrow to reach the ground is approximately \( t \approx 14.94 \) seconds.