Eric is asked to kick a ball from one end of the soccer field to the other end, which are 100 m apart. If the ball leaves the ground at an angle \( 60^{\circ} \) to the ground and takes 5 s to reach the other end, calculate the following: 1.3.1 The velocity at which the ball leaves the ground .3 .2 The maximum height the ball reaches before it makes its way back to the ground

To solve Eric's projectile motion problem, we'll analyze the motion in both the horizontal and vertical directions separately. ### Given: - **Horizontal Distance (\( R \))**: 100 meters - **Angle of Projection (\( \theta \))**: \( 60^{\circ} \) - **Time of Flight (\( T \))**: 5 seconds - **Acceleration due to Gravity (\( g \))**: \( 9.81 \, \text{m/s}^2 \) ### 1.3.1 **Velocity at Which the Ball Leaves the Ground (\( v_0 \))** **Step 1: Analyze Horizontal Motion** The horizontal component of the initial velocity (\( v_{0x} \)) is given by: \[ v_{0x} = v_0 \cos(\theta) \] The horizontal distance covered is: \[ R = v_{0x} \times T \] \[ 100 \, \text{m} = v_0 \cos(60^{\circ}) \times 5 \, \text{s} \] \[ 100 = v_0 \times 0.5 \times 5 \] \[ 100 = 2.5 \, v_0 \] \[ v_0 = \frac{100}{2.5} = 40 \, \text{m/s} \] **Answer:** The ball leaves the ground with an initial velocity of **40  m/s**. --- ### 1.3.2 **Maximum Height Reached (\( H \))** **Step 2: Analyze Vertical Motion** The vertical component of the initial velocity (\( v_{0y} \)) is: \[ v_{0y} = v_0 \sin(\theta) = 40 \times \sin(60^{\circ}) = 40 \times \frac{\sqrt{3}}{2} \approx 34.64 \, \text{m/s} \] The maximum height \( H \) can be calculated using the formula: \[ H = \frac{v_{0y}^2}{2g} \] \[ H = \frac{(34.64)^2}{2 \times 9.81} \approx \frac{1199.3}{19.62} \approx 61.1 \, \text{meters} \] **Answer:** The maximum height the ball reaches is approximately **61 meters**. --- ### **Summary of Answers:** 1. **Initial Velocity (\( v_0 \))**: 40 m/s 2. **Maximum Height (\( H \))**: ≈61 meters