Let's break down the problem step by step: --- ## **Given Data** - Weight of golf ball (\( W \)): **.1 lb** - Angle of launch (\( \theta \)): **30°** - Range (\( R \)): **500 ft** - Time of contact (\( \Delta t \)): **.5 ms = .5 \times 10^{-3} \) s** --- ## **Step 1: Find Initial Velocity (\( v_ \))** The range for projectile motion is: \[ R = \frac{v_^2 \sin 2\theta}{g} \] Where: - \( R = 500 \) ft - \( \theta = 30^\circ \) - \( g = 32.2 \) ft/s² (standard gravity) First, calculate \( \sin 2\theta \): \[ \sin(2 \times 30^\circ) = \sin(60^\circ) = .866 \] Plug into the range formula and solve for \( v_ \): \[ 500 = \frac{v_^2 \cdot .866}{32.2} \] \[ v_^2 = \frac{500 \times 32.2}{.866} \] \[ v_^2 = \frac{16100}{.866} \approx 18593.77 \] \[ v_ = \sqrt{18593.77} \approx 136.34 \text{ ft/s} \] --- ## **Step 2: Find Change in Momentum (\( \Delta p \))** First, find the **mass** (\( m \)): \[ W = mg \implies m = \frac{W}{g} = \frac{.1}{32.2} = .00311 \text{ slugs} \] The change in momentum (assuming the initial velocity is ): \[ \Delta p = m v_ \] \[ \Delta p = .00311 \times 136.34 \approx .4245 \text{ slug-ft/s} \] --- ## **Step 3: Average Impulsive Force (\( F_{avg} \))** Impulse is also: \[ \text{Impulse} = F_{avg} \Delta t = \Delta p \] \[ F_{avg} = \frac{\Delta p}{\Delta t} \] \[ \Delta t = .5\, \text{ms} = .0005\, \text{s} \] \[ F_{avg} = \frac{.4245}{.0005} = 849 \text{ lb} \] --- ## **Step 4: Final Answer** \[ \boxed{F_{avg} = 849 \text{ lb}} \] --- ### **Summary of Steps** 1. Used range equation to find initial velocity. 2. Converted weight to mass. 3. Used impulse-momentum theorem to find average force. --- **The average impulsive force imparted by the club to the ball is approximately \( 849 \) lb.**

# Step-by-Step Solution to Impulsive Force of a Golf Club on a Ball ## Given Data - **Weight of Golf Ball (\( W \))**: 0.1 lb - **Angle of Launch (\( \theta \))**: 30° - **Range (\( R \))**: 500 ft - **Time of Contact (\( \Delta t \))**: 0.5 ms = \( 0.5 \times 10^{-3} \) s ## Step 1: Find Initial Velocity (\( v_ \)) The range formula for projectile motion is given by: \[ R = \frac{v^2 \sin(2\theta)}{g} \] Where: - \( g \) = 32.2 ft/s² (acceleration due to gravity) ### Calculate \( \sin(2\theta) \) \[ \sin(2 \times 30^\circ) = \sin(60^\circ) = 0.866 \] ### Solve for \( v_ \) Substituting known values into the range equation: \[ 500 = \frac{v^2 \cdot 0.866}{32.2} \] Rearranging gives: \[ v^2 = \frac{500 \times 32.2}{0.866} \] Calculating: \[ v^2 \approx \frac{16100}{0.866} \approx 18593.77 \] \[ v_ \approx \sqrt{18593.77} \approx 136.34 \text{ ft/s} \] ## Step 2: Find Change in Momentum (\( \Delta p \)) ### Convert Weight to Mass Using the relation \( W = mg \): \[ m = \frac{W}{g} = \frac{0.1}{32.2} \approx 0.00311 \text{ slugs} \] ### Calculate Change in Momentum Assuming the initial velocity is 0, the change in momentum is: \[ \Delta p = mv_ \approx 0.00311 \times 136.34 \approx 0.4245 \text{ slug-ft/s} \] ## Step 3: Average Impulsive Force (\( F_{avg} \)) Using the impulse-momentum theorem: \[ \text{Impulse} = F_{avg} \Delta t = \Delta p \] ### Solve for \( F_{avg} \) \[ F_{avg} = \frac{\Delta p}{\Delta t} \] Where \( \Delta t = 0.5 \, \text{ms} = 0.0005 \, \text{s} \): \[ F_{avg} = \frac{0.4245}{0.0005} \approx 849 \text{ lb} \] ## Final Answer \[ \boxed{F_{avg} = 849 \text{ lb}} \] --- ### Summary of Steps 1. Calculated initial velocity using the range formula. 2. Converted weight to mass to find the change in momentum. 3. Applied the impulse-momentum theorem to determine the average force. **The average impulsive force imparted by the club to the ball is approximately \( 849 \) lb.**