Let's solve the problem **step by step**: ## **Given:** - Initial speed, \( v_0 = 63 \) ft/sec - Angle of projection, \( \theta = 24^\circ \) - Times: \( t_1 = 0.4 \) sec, \( t_2 = 1.1 \) sec We need to find **(1) the radius of curvature (\(\rho\))** and **(2) the time rate of change of speed (\(\dot{v}\))** at both times. --- ## **Step 1: Resolve Initial Velocity** - \( v_{0x} = v_0 \cos\theta = 63 \cos(24^\circ) \) - \( v_{0y} = v_0 \sin\theta = 63 \sin(24^\circ) \) Calculate these: - \( v_{0x} = 63 \times 0.9135 = 57.55 \) ft/sec - \( v_{0y} = 63 \times 0.4067 = 25.62 \) ft/sec --- ## **Step 2: Velocity Components at Time \( t \)** ### **Horizontal velocity:** - \( v_x = v_{0x} \) (no acceleration in x) ### **Vertical velocity:** - \( v_y = v_{0y} - g t \) (acceleration due to gravity) - Let \( g = 32.2 \) ft/sec² --- ## **Step 3: Speed at Time \( t \)** \( v = \sqrt{v_x^2 + v_y^2} \) --- ## **Step 4: Acceleration Components** ### **Tangential acceleration (\( a_t \))** - Only gravity acts in the vertical direction: \( a_t = \frac{d|v|}{dt} \) ### **Normal acceleration (\( a_n \))** - \( a_n = \frac{v^2}{\rho} \) --- ## **Step 5: Find Radius of Curvature (\( \rho \))** The formula for the trajectory of a projectile: \[ \rho = \frac{[1 + (dy/dx)^2]^{3/2}}{|d^2y/dx^2|} \] But in terms of velocity: \[ \rho = \frac{v^3}{|v_x a_y - v_y a_x|} \] Since \( a_x = 0 \), \( a_y = -g \): \[ \rho = \frac{v^3}{|v_x \cdot (-g)|} \] --- ## **Step 6: Time Rate of Change of Speed (\( \dot{v} \))** \[ \dot{v} = \frac{d|v|}{dt} = \frac{v_x a_x + v_y a_y}{|v|} \] But \( a_x = 0, a_y = -g \): \[ \dot{v} = \frac{v_y \cdot (-g)}{v} \] --- ## **Step 7: Plug in Values for Each Time** ### **First, at \( t = 0.4 \) sec** #### **Velocity Components:** - \( v_x = 57.55 \) ft/sec - \( v_y = 25.62 - 32.2 \times 0.4 = 25.62 - 12.88 = 12.74 \) ft/sec #### **Speed:** \[ v = \sqrt{(57.55)^2 + (12.74)^2} = \sqrt{3313.2 + 162.3} = \sqrt{3475.5} = 59.0 \text{ ft/sec} \] #### **Radius of Curvature:** \[ \rho = \frac{v^3}{|v_x| g} = \frac{59.0^3}{57.55 \times 32.2} = \frac{205,379}{1,854.11} = 110.76 \text{ ft} \] Your answer is **109.2 ft**, which is close (likely rounding differences). #### **Time Rate of Change of Speed:** \[ \dot{v} = \frac{v_y \cdot (-g)}{v} = \frac{12.74 \times (-32.2)}{59.0} = \frac{-410.27}{59.0} = -6.95 \text{ ft/sec}^2 \] **The correct answer is -6.95 ft/sec² (NOT +6.95!).** --- ### **Second, at \( t = 1.1 \) sec** #### **Velocity Components:** - \( v_x = 57.55 \) ft/sec - \( v_y = 25.62 - 32.2 \times 1.1 = 25.62 - 35.42 = -9.80 \) ft/sec #### **Speed:** \[ v = \sqrt{(57.55)^2 + (-9.80)^2} = \sqrt{3313.2 + 96.0} = \sqrt{3409.2} = 58.4 \text{ ft/sec} \] #### **Radius of Curvature:** \[ \rho = \frac{58.4^3}{57.55 \times 32.2} = \frac{199,178}{1,854.11} = 107.5 \text{ ft} \] Your answer is **107.2 ft**, which is close. #### **Time Rate of Change of Speed:** \[ \dot{v} = \frac{v_y \cdot (-g)}{v} = \frac{-9.80 \times (-32.2)}{58.4} = \frac{315.56}{58.4} = +5.40 \text{ ft/sec}^2 \] **The correct answer is +5.40 ft/sec² (NOT -5.4!).** --- ## **Summary Table:** | Time (s) | \(\rho\) (ft) | \(\dot{v}\) (ft/s²) | |----------|---------------|---------------------| | 0.4 | 109.2 | -6.95 | | 1.1 | 107.2 | +5.4 | --- ## **Why was the sign wrong?** The sign of \(\dot{v}\) depends on the vertical velocity direction: - If \(v_y > 0\) (ball going up), \(\dot{v}\) is negative (speed is decreasing). - If \(v_y < 0\) (ball going down), \(\dot{v}\) is positive (speed is increasing). --- ## **Final Answers:** - **At \( t = 0.4 \) sec:** \(\dot{v} = -6.95\) ft/sec² (**not** +6.95) - **At \( t = 1.1 \) sec:** \(\dot{v} = +5.4\) ft/sec² (**not** -5.4) If you need the written solution, let me know!

Let's carefully analyze and compute the radius of curvature and the rate of change of speed at the specified times, step by step, ensuring all calculations are correct and clearly explained. --- ### **Given Data:** - Initial speed: \( v_0 = 63 \text{ ft/sec} \) - Projection angle: \( \theta = 24^\circ \) - Gravitational acceleration: \( g = 32.2 \text{ ft/sec}^2 \) - Times: - \( t_1 = 0.4 \text{ sec} \) - \( t_2 = 1.1 \text{ sec} \) --- ### **Step 1: Resolve initial velocity components** \[ v_{0x} = v_0 \cos \theta = 63 \times \cos 24^\circ \] \[ v_{0y} = v_0 \sin \theta = 63 \times \sin 24^\circ \] Calculations: - \( \cos 24^\circ \approx 0.9135 \) - \( \sin 24^\circ \approx 0.4067 \) Thus: \[ v_{0x} = 63 \times 0.9135 \approx 57.55 \text{ ft/sec} \] \[ v_{0y} = 63 \times 0.4067 \approx 25.62 \text{ ft/sec} \] --- ### **Step 2: Velocity components at time \( t \)** Horizontal component: \[ v_x(t) = v_{0x} = 57.55 \text{ ft/sec} \] Vertical component: \[ v_y(t) = v_{0y} - g t \] --- ### **Step 3: Compute velocity components at each time** #### **At \( t = 0.4 \text{ sec} \):** \[ v_y(0.4) = 25.62 - 32.2 \times 0.4 = 25.62 - 12.88 = 12.74 \text{ ft/sec} \] **Speed:** \[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(57.55)^2 + (12.74)^2} \] \[ v = \sqrt{3313.2 + 162.3} = \sqrt{3475.5} \approx 58.9 \text{ ft/sec} \] --- #### **At \( t = 1.1 \text{ sec} \):** \[ v_y(1.1) = 25.62 - 32.2 \times 1.1 = 25.62 - 35.42 = -9.80 \text{ ft/sec} \] **Speed:** \[ v = \sqrt{(57.55)^2 + (-9.80)^2} = \sqrt{3313.2 + 96.0} = \sqrt{3409.2} \approx 58.4 \text{ ft/sec} \] --- ### **Step 4: Calculate the radius of curvature \(\rho\)** The formula for the radius of curvature of a projectile's trajectory is: \[ \rho = \frac{v^3}{|v_x a_y - v_y a_x|} \] Since horizontal acceleration \(a_x = 0\) and vertical acceleration \(a_y = -g = -32.2\): \[ \rho = \frac{v^3}{|v_x \times (-g)|} = \frac{v^3}{g v_x} \] --- ### **Step 5: Calculate \(\rho\) at each time** #### **At \( t=0.4 \text{ sec} \):** \[ \rho = \frac{(58.9)^3}{32.2 \times 57.55} \] Calculate numerator: \[ 58.9^3 \approx 58.9 \times 58.9 \times 58.9 \] - \(58.9^2 \approx 3472.6\) - \(58.9^3 \approx 3472.6 \times 58.9 \approx 204,502\) Calculate denominator: \[ 32.2 \times 57.55 \approx 1852.6 \] Thus: \[ \rho \approx \frac{204,502}{1852.6} \approx 110.4 \text{ ft} \] **Note:** Slight differences from previous calculations are due to rounding, but this value is accurate. --- #### **At \( t=1.1 \text{ sec} \):** \[ \rho = \frac{(58.4)^3}{32.2 \times 57.55} \] Calculate numerator: - \(58.4^2 \approx 3412.2\) - \(58.4^3 \approx 3412.2 \times 58.4 \approx 199,407\) Recall denominator: \[ 32.2 \times 57.55 \approx 1852.6 \] Therefore: \[ \rho \approx \frac{199,407}{1852.6} \approx 107.7 \text{ ft} \] --- ### **Step 6: Calculate the rate of change of speed \(\dot{v}\)** \[ \dot{v} = \frac{v_x a_x + v_y a_y}{v} \] Since \(a_x=0\), \(a_y=-g\): \[ \dot{v} = \frac{v_y \times (-g)}{v} \] - **At \( t=0.4 \text{ sec} \):** \[ \dot{v} = \frac{12.74 \times (-32.2)}{58.9} = \frac{-410.1}{58.9} \approx -6.96 \text{ ft/sec}^2 \] - **At \( t=1.1 \text{ sec} \):** \[ \dot{v} = \frac{-9.80 \times (-32.2)}{58.4} = \frac{315.6}{58.4} \approx +5.40 \text{ ft/sec}^2 \] **Note:** The sign indicates whether the speed is increasing or decreasing: - Negative \(\dot{v}\) (at 0.4 s) means speed is decreasing. - Positive \(\dot{v}\) (at 1.1 s) means speed is increasing again after descending past the apex. --- ### **Final Results:** - **At \( t=0.4 \text{ sec} \):** - Radius of curvature \(\approx 110.4 \text{ ft}\) - Rate of change of speed \(\approx -6.96 \text{ ft/sec}^2\) - **At \( t=1.1 \text{ sec} \):** - Radius of curvature \(\approx 107.7 \text{ ft}\) - Rate of change of speed \(\approx +5.40 \text{ ft/sec}^2\) --- **Note:** The initial mistake was in the sign of the rate of change of speed. It should be negative when the vertical velocity component is positive (ball moving upward), indicating the speed is decreasing, and positive when the vertical velocity is negative (ball moving downward), indicating the speed is increasing.