To calculate the range of a projectile, we need to use the standard kinematic equation for the horizontal range:

$R = \frac{v_0^2 \sin(2\theta)}{g}$

Where:

• $v_0$ is the initial velocity of the projectile.
• $\theta$ is the angle of projection above the horizontal.
• $g$ is the acceleration due to gravity, which is approximately $9.81 \, \text{m/s}^2$.

Step 1: Convert Initial Velocity to Meters per Second

The initial velocity $v_0$ is given in miles per hour (mph). To convert it to meters per second (m/s):

$1 \, \text{mile} = 1609.34 \, \text{meters}$ $1 \, \text{hour} = 3600 \, \text{seconds}$

So, the conversion factor is: $\frac{1609.34}{3600} \approx 0.44704 \, \text{m/s per mph}$

Now, convert the given initial velocity: $v_0 = 19.45 \, \text{mph} \times 0.44704 \, \text{m/s per mph} \approx 8.693 \, \text{m/s}$

Step 2: Plug Values into the Range Formula

Now, use the formula for range, where:

• $v_0 = 8.693 \, \text{m/s}$
• $\theta = 37.5^\circ$
• $g = 9.81 \, \text{m/s}^2$

First, calculate $\sin(2\theta)$: $2\theta = 2 \times 37.5^\circ = 75^\circ$ $\sin(75^\circ) \approx 0.9659$

Now, substitute everything into the range formula:

$R = \frac{(8.693 \, \text{m/s})^2 \times 0.9659}{9.81 \, \text{m/s}^2}$

Simplify:

$R = \frac{75.5489 \times 0.9659}{9.81} \approx \frac{72.96}{9.81} \approx 7.44 \, \text{meters}$

Final Answer:

The range of the projectile is approximately 7.44 meters.

Would you like more details or have any questions?

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Here are 5 questions that extend this concept:

1. How would the range change if the angle was increased to 45 degrees?
2. What would happen to the range if the projectile was launched from a height instead of level ground?
3. How would air resistance affect the range of the projectile?
4. How can we determine the maximum height of the projectile in this scenario?
5. What is the total time the projectile spends in the air?

Tip: Always ensure unit consistency when solving physics problems, especially when converting between imperial and metric systems.