Physics Solution: Question 13

Solution to Question 13

1. Analyzing the Velocity-Time Graph

The problem provides the velocity-time graph of a ball moving vertically under the influence of gravity and air resistance. To solve for the impact speed, we first need to determine the velocity scale $v_0$ by analyzing the motion at the highest point.

Key Concept: At the highest point of the trajectory ($t=2$ s), the instantaneous velocity is zero ($v=0$).

Since air resistance is velocity-dependent ($F_{drag} \propto v$ or $v^2$), the drag force is zero when $v=0$.
Therefore, at the exact instant $t=2$ s, the only force acting on the ball is gravity.
The acceleration corresponds to the slope of the tangent to the $v-t$ graph. Thus, at $v=0$, the slope must be equal to the acceleration due to gravity, $-g$.

Figure: $v-t$ graph showing the smooth curve tangent to the line with slope $-g$ at $t=2$.

2. Calculating $v_0$

We use the slope of the red tangent line at $t=2$ to find $v_0$.

From the grid:

At $t=2$, $v=0$.
Moving left by $\Delta t = 0.5$ s (one grid unit), the tangent line rises to $v = 0.5 v_0$ (one vertical grid unit).

Calculating the magnitude of the slope: $$ | \text{slope} | = \left| \frac{\Delta v}{\Delta t} \right| = \frac{0.5 v_0}{0.5} = v_0 $$

Since the slope at the highest point is $g$: $$ v_0 = g $$

            Using $g = 10 \, \mathrm{m/s^2}$, we find:
            $$ v_0 = 10 \, \mathrm{m/s} $$
        

3. Determining Impact Speed

The ball strikes the ground at its terminal velocity, which is the asymptotic value on the graph.

Observing the $y$-axis:

The curve flattens out and ends at the horizontal line marked $-2.5 v_0$.
This indicates the final velocity is $-2.5 v_0$.

The speed is the magnitude of the velocity: $$ \text{Speed} = | -2.5 v_0 | = 2.5 \times 10 \, \mathrm{m/s} $$

$$ \text{Speed} = 25 \, \mathrm{m/s} $$

Final Answer: The ball strikes the ground with a speed of $25 \, \mathrm{m/s}$.