Let's break down the problem step by step: **Given:** - **Horizontal range per hop (\( R \))**: \( 10 \) m - **Launch angle (\( \theta \))**: \( 20^\circ \) - **Dingoes' speed**: \( 50 \) km/hr We want to determine the kangaroo's speed and see if it is greater than that of the dingoes. --- ### **Step 1: Find the Kangaroo's Take-off Speed** The **range** of a projectile (ignoring air resistance) is given by: \[ R = \frac{v^2 \sin 2\theta}{g} \] where: - \( v \) = take-off speed - \( \theta \) = launch angle - \( g \) = acceleration due to gravity (\( 9.8\, \text{m/s}^2 \)) - \( R \) = horizontal distance per hop (\( 10\, \text{m} \)) Plug in the values: \[ R = 10\, \text{m},\quad \theta = 20^\circ \] Calculate \( \sin 2\theta = \sin 40^\circ \). \[ \sin 40^\circ \approx 0.6428 \] So, \[ 10 = \frac{v^2 \times 0.6428}{9.8} \] \[ 10 \times 9.8 = v^2 \times 0.6428 \] \[ 98 = v^2 \times 0.6428 \] \[ v^2 = \frac{98}{0.6428} \approx 152.46 \] \[ v = \sqrt{152.46} \approx 12.35\, \text{m/s} \] --- ### **Step 2: Find Average Horizontal Speed of Kangaroo** The horizontal speed is: \[ v_x = v \cos\theta \] \[ = 12.35 \times \cos 20^\circ \] \[ \cos 20^\circ \approx 0.9397 \] \[ v_x = 12.35 \times 0.9397 \approx 11.59\, \text{m/s} \] --- ### **Step 3: Find the Speed of Dingoes in m/s** \[ 50\, \text{km/hr} = 50 \times \frac{1000}{3600} = 13.89\, \text{m/s} \] --- ### **Step 4: Compare the Speeds** - Kangaroo's horizontal speed: **11.59 m/s** - Dingoes' speed: **13.89 m/s** --- ### **Conclusion** **The kangaroo's speed (11.59 m/s) is less than the dingoes' speed (13.89 m/s).** **Therefore, at this speed, the kangaroo would NOT be able to escape from the pack of dingoes.** --- ### **Final Answer** > **No, at this take-off speed and angle, the kangaroo would not be able to escape from the dingoes moving at 50 km/hr, because its horizontal speed (11.59 m/s) is less than the dingoes' speed (13.89 m/s).**