Solution to Question 11
Problem Analysis: We are observing a projectile’s trajectory compared to the direct “Line of Sight” (aim). The vertical deviation from the line of sight due to gravity is a powerful tool to solve this without finding the angle or initial velocity explicitly.
1. Analyze the Graph Data
From the problem graph, we can extract coordinates:
- Aim Point: At wall distance $x = 20$ m, height $y = 50$ m.
- Line of Sight Equation: Since it passes through $(0,0)$ and $(20,50)$, the slope is $m = 50/20 = 2.5$. Thus, $y_{aim} = 2.5x$.
- The Circular Dot: The dot is located at $x = 10$. Looking at the grid, the dot’s height is $y = 20$.
2. Calculate Deviation at the Dot
At $x=10$, the height of the line of sight is:
$$y_{aim}(10) = 2.5 \times 10 = 25 \text{ m}$$The actual height of the bullet is $20$ m. The vertical drop due to gravity ($h$) is:
$$h = y_{aim} – y_{actual} = 25 – 20 = 5 \text{ m}$$3. Determine Time of Flight
The vertical drop from the line of sight is given by free fall: $h = \frac{1}{2}gt^2$.
Using $g = 10 \text{ m/s}^2$:
$$5 = \frac{1}{2}(10)t^2 \implies 5 = 5t^2 \implies t = 1 \text{ s}$$So, it takes 1 second to travel 10 meters horizontally.
4. Calculate Impact on Wall
The wall is at $x=20$ m. Since horizontal velocity is constant, the time to reach the wall is double the time to reach $x=10$.
$$t_{wall} = 2 \times 1 \text{ s} = 2 \text{ s}$$Now, calculate the vertical drop at the wall:
$$H_{drop} = \frac{1}{2} g t_{wall}^2 = \frac{1}{2}(10)(2)^2 = 20 \text{ m}$$The bullet was aimed at the 50 m mark. It drops 20 m below the aim.
$$y_{hit} = 50 \text{ m} – 20 \text{ m} = 30 \text{ m}$$