Question

A bullet is fired in a horizontal direction with a muzzle velocity of 300m\s in the absence of air resistance how far will it have dropped in traveling a horizontal distance for : 20m, 40m, 60m

Answer 1

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:

1. **For \( x = 20 \) m:**
\[
t = \frac{20 \, \text{m}}{300 \, \text{m/s}} = \frac{1}{15} \, \text{s} \approx 0.06667 \, \text{s}
\]
\[
y = \frac{1}{2} \cdot 9.81 \, \text{m/s}^2 \cdot \left(\frac{1}{15}\right)^2 \approx \frac{1}{2} \cdot 9.81 \cdot 0.0004444 \approx 0.00218 \, \text{m} \approx 2.18 \, \text{mm}
\]

2. **For \( x = 40 \) m:**
\[
t = \frac{40 \, \text{m}}{300 \, \text{m/s}} = \frac{2}{15} \, \text{s} \approx 0.13333 \, \text{s}
\]
\[
y = \frac{1}{2} \cdot 9.81 \, \text{m/s}^2 \cdot \left(\frac{2}{15}\right)^2 = \frac{1}{2} \cdot 9.81 \cdot 0.011111 \approx 0.05455 \, \text{m} \approx 54.55 \, \text{mm}
\]

3. **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**