Question

An arrow is shot with an initial velocity of 63. m/s at an angle of 45° above the horizontal. What is the horizontal distance traveled during the flight?

Answer 1

To find the horizontal distance traveled by the arrow, we can use the equations of projectile motion.

1. **Determine the initial velocity components:**
The initial velocity \( v_0 = 63.0 \, \text{m/s} \) is at an angle of \( \theta = 45^\circ \).

The horizontal (x) and vertical (y) components of the initial velocity can be calculated using:
\[
v_{0x} = v_0 \cdot \cos(\theta)
\]
\[
v_{0y} = v_0 \cdot \sin(\theta)
\]

Given \(\theta = 45^\circ\):
\[
v_{0x} = 63.0 \cdot \cos(45^\circ) = 63.0 \cdot \frac{\sqrt{2}}{2} \approx 44.6 \, \text{m/s}
\]
\[
v_{0y} = 63.0 \cdot \sin(45^\circ) = 63.0 \cdot \frac{\sqrt{2}}{2} \approx 44.6 \, \text{m/s}
\]

2. **Determine the time of flight:**
The time of flight \( T \) can be found from the vertical motion. The total time of flight for a projectile launched and landing at the same height can be calculated using:
\[
T = \frac{2 v_{0y}}{g}
\]
where \( g = 9.81 \, \text{m/s}^2 \) (the acceleration due to gravity).

Plugging in the values:
\[
T = \frac{2 \cdot 44.6}{9.81} \approx \frac{89.2}{9.81} \approx 9.09 \, \text{s}
\]

3. **Calculate the horizontal distance:**
The horizontal distance \( R \) (range) can be calculated using:
\[
R = v_{0x} \cdot T
\]

Substituting the values we found:
\[
R \approx 44.6 \, \text{m/s} \cdot 9.09 \, \text{s} \approx 405.78 \, \text{m}
\]

Therefore, the horizontal distance traveled during the flight is approximately **405.78 meters**.