ball is thrown upward at \( 20 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) off the top of a building that is 12 m high. What is the height of the ball above the ground at its highest point? How long does it take the ball to reach its highest point? What is the total time taken for the ball to reach the ground after leaving the person's hand?

Let's analyze the problem step-by-step using the kinematic equations of motion. We'll assume the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) acting downward. ### Given: - **Initial velocity (\( u \))**: \( 20 \, \text{m/s} \) (upward) - **Height of the building (\( h_{\text{building}} \))**: \( 12 \, \text{m} \) - **Acceleration due to gravity (\( a \))**: \( -9.8 \, \text{m/s}^2 \) (negative since it's downward) --- ### 1. **Height of the Ball at Its Highest Point** At the highest point, the vertical velocity (\( v \)) becomes zero. We can use the following kinematic equation to find the displacement (\( s \)) from the initial position (top of the building) to the highest point: \[ v^2 = u^2 + 2a s \] Plugging in the known values: \[ 0 = (20)^2 + 2(-9.8)s \\ 0 = 400 - 19.6s \\ 19.6s = 400 \\ s = \frac{400}{19.6} \approx 20.41 \, \text{m} \] **Total height above the ground:** \[ h_{\text{total}} = h_{\text{building}} + s = 12 \, \text{m} + 20.41 \, \text{m} \approx 32.41 \, \text{m} \] **Rounded Answer:** Approximately **32.4 meters** above the ground. --- ### 2. **Time to Reach the Highest Point** We'll use the equation: \[ v = u + a t \] Setting \( v = 0 \) at the highest point: \[ 0 = 20 - 9.8t \\ 9.8t = 20 \\ t = \frac{20}{9.8} \approx 2.04 \, \text{seconds} \] **Rounded Answer:** Approximately **2.04 seconds** to reach the highest point. --- ### 3. **Total Time to Reach the Ground** To find the total time from the moment the ball is thrown until it hits the ground, we'll consider the entire motion: upwards to the highest point and then downwards to the ground. Use the displacement equation, where the final displacement (\( s \)) is \( -12 \, \text{m} \) (since it ends 12 meters below the starting point): \[ s = ut + \frac{1}{2} a t^2 \\ -12 = 20t - 4.9t^2 \\ 4.9t^2 - 20t - 12 = 0 \] Solving this quadratic equation for \( t \): \[ t = \frac{20 \pm \sqrt{(20)^2 - 4 \cdot 4.9 \cdot (-12)}}{2 \cdot 4.9} \\ t = \frac{20 \pm \sqrt{400 + 235.2}}{9.8} \\ t = \frac{20 \pm \sqrt{635.2}}{9.8} \\ t = \frac{20 \pm 25.2}{9.8} \] Discarding the negative root (since time cannot be negative): \[ t = \frac{45.2}{9.8} \approx 4.61 \, \text{seconds} \] **Rounded Answer:** Approximately **4.61 seconds** to reach the ground. --- ### **Summary of Answers:** 1. **Highest Point Height:** Approximately **32.4 meters** above the ground. 2. **Time to Reach Highest Point:** Approximately **2.04 seconds**. 3. **Total Time to Reach the Ground:** Approximately **4.61 seconds**.