Question
If an arrow is fired from a bow with a perfectly horizontal velocity of 22.0 m/s and the arrow was 4.0 m above the ground when the it was released, how far will the arrow fly horizontally before it strikes the ground?
Group of answer choices
39.1 m
15.0 m
16.1 m
19.9 m
Group of answer choices
39.1 m
15.0 m
16.1 m
19.9 m
Answers
GPT-4o mini
To solve this problem, we need to calculate the time it takes for the arrow to fall 4.0 meters vertically and then use that time to find out how far it travels horizontally.
### Step 1: Calculate the time of flight
The arrow falls under the influence of gravity. We can use the following kinematic equation for vertical motion:
\[
d = v_i \cdot t + \frac{1}{2} g t^2
\]
Where:
- \( d \) is the vertical distance fallen (4.0 m),
- \( v_i \) is the initial vertical velocity (0 m/s, since it's fired horizontally),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( t \) is the time in seconds.
Since the initial velocity in the vertical direction is 0, the equation simplifies to:
\[
d = \frac{1}{2} g t^2
\]
Plugging in the values, we have:
\[
4.0 = \frac{1}{2} (9.81) t^2
\]
\[
4.0 = 4.905 t^2
\]
Now solving for \( t^2 \):
\[
t^2 = \frac{4.0}{4.905} \approx 0.815
\]
Taking the square root:
\[
t \approx \sqrt{0.815} \approx 0.903 \, \text{s}
\]
### Step 2: Calculate the horizontal distance traveled
Now that we have the time of flight, we can find the horizontal distance traveled using the formula:
\[
d = v \cdot t
\]
Where:
- \( d \) is the horizontal distance,
- \( v \) is the horizontal velocity (22.0 m/s),
- \( t \) is the time of flight we just calculated.
Substituting in the values:
\[
d = 22.0 \, \text{m/s} \cdot 0.903 \, \text{s} \approx 19.9 \, \text{m}
\]
### Conclusion
The arrow will fly approximately **19.9 m** horizontally before it strikes the ground.
Thus, the correct answer is **19.9 m**.
### Step 1: Calculate the time of flight
The arrow falls under the influence of gravity. We can use the following kinematic equation for vertical motion:
\[
d = v_i \cdot t + \frac{1}{2} g t^2
\]
Where:
- \( d \) is the vertical distance fallen (4.0 m),
- \( v_i \) is the initial vertical velocity (0 m/s, since it's fired horizontally),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( t \) is the time in seconds.
Since the initial velocity in the vertical direction is 0, the equation simplifies to:
\[
d = \frac{1}{2} g t^2
\]
Plugging in the values, we have:
\[
4.0 = \frac{1}{2} (9.81) t^2
\]
\[
4.0 = 4.905 t^2
\]
Now solving for \( t^2 \):
\[
t^2 = \frac{4.0}{4.905} \approx 0.815
\]
Taking the square root:
\[
t \approx \sqrt{0.815} \approx 0.903 \, \text{s}
\]
### Step 2: Calculate the horizontal distance traveled
Now that we have the time of flight, we can find the horizontal distance traveled using the formula:
\[
d = v \cdot t
\]
Where:
- \( d \) is the horizontal distance,
- \( v \) is the horizontal velocity (22.0 m/s),
- \( t \) is the time of flight we just calculated.
Substituting in the values:
\[
d = 22.0 \, \text{m/s} \cdot 0.903 \, \text{s} \approx 19.9 \, \text{m}
\]
### Conclusion
The arrow will fly approximately **19.9 m** horizontally before it strikes the ground.
Thus, the correct answer is **19.9 m**.