To calculate how far a bullet falls vertically while traveling horizontally, we can use the equations of motion. In the absence of air resistance, the vertical drop \( y \) can be determined by the time \( t \) it takes to travel a certain horizontal distance \( x \).
The relationship between horizontal distance and time is given by:
\[ t = \frac{x}{v} \]
where:
- \( v = 300 \) m/s (muzzle velocity)
- \( x \) is the horizontal distance traveled.
The vertical drop \( y \) due to gravity can be calculated using the equation:
\[ y = \frac{1}{2} g t^2 \]
where:
- \( g = 9.81 \) m/s² (acceleration due to gravity)
- \( t \) is the time calculated from the horizontal distance.
Let's calculate the drop for each horizontal distance:
-
For \( x = 20 \) m: \[ t = \frac{20 , \text{m}}{300 , \text{m/s}} = \frac{1}{15} , \text{s} \approx 0.06667 , \text{s} \] \[ y = \frac{1}{2} \cdot 9.81 , \text{m/s}^2 \cdot \left(\frac{1}{15}\right)^2 \approx \frac{1}{2} \cdot 9.81 \cdot 0.0004444 \approx 0.00218 , \text{m} \approx 2.18 , \text{mm} \]
-
For \( x = 40 \) m: \[ t = \frac{40 , \text{m}}{300 , \text{m/s}} = \frac{2}{15} , \text{s} \approx 0.13333 , \text{s} \] \[ y = \frac{1}{2} \cdot 9.81 , \text{m/s}^2 \cdot \left(\frac{2}{15}\right)^2 = \frac{1}{2} \cdot 9.81 \cdot 0.011111 \approx 0.05455 , \text{m} \approx 54.55 , \text{mm} \]
-
For \( x = 60 \) m: \[ t = \frac{60 , \text{m}}{300 , \text{m/s}} = \frac{1}{5} , \text{s} = 0.2 , \text{s} \] \[ y = \frac{1}{2} \cdot 9.81 , \text{m/s}^2 \cdot (0.2)^2 = \frac{1}{2} \cdot 9.81 \cdot 0.04 \approx 0.1962 , \text{m} \approx 196.2 , \text{mm} \]
Summary of vertical drop:
- At 20 m: 2.18 mm
- At 40 m: 54.55 mm
- At 60 m: 196.2 mm
11 answers