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?

To solve this problem, we can break it down step-by-step! **Height at the Highest Point:** The maximum height can be calculated using the formula \( h = v^2 / (2g) \), where \( v \) is the initial velocity (20 m/s), and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). The time to reach the highest point is when the velocity becomes 0. 1. **Height at the highest point above the building:** \[ \text{Additional height} = \frac{(20)^2}{2 \times 9.81} \approx 20.4 \text{ m} \] Thus, total height above the ground: \[ \text{Total Height} = 12 \text{ m (height of the building)} + 20.4 \text{ m} \approx 32.4 \text{ m} \] **Time to Reach Highest Point:** Using the kinematic equation \( v = u + at \), where \( v \) is the final velocity (0 m/s), \( u \) is the initial velocity (20 m/s), \( a \) is the acceleration (-9.81 m/s² due to gravity), we get: \[ 0 = 20 - 9.81t \] \[ t \approx \frac{20}{9.81} \approx 2.04 \text{ seconds} \] **Total Time Taken to Reach the Ground:** The total time can be found by calculating the time to ascend to the highest point and then the time to descend from that height back to the ground. 1. **Time to descend from highest point to ground:** Use the formula for distance \( d = 0.5 g t^2 \) starting from the height of 32.4 m: \[ 32.4 = 0.5 \times 9.81 \times t^2 \] \[ t^2 \approx \frac{32.4 \times 2}{9.81} \] \[ t^2 \approx 6.61 \] \[ t \approx 2.57 \text{ seconds} \] 2. **Total time = time to rise + time to fall:** \[ \text{Total Time} = 2.04 + 2.57 \approx 4.61 \text{ seconds} \] So, the height of the ball at its highest point is approximately 32.4 m above the ground, it takes about 2.04 seconds to reach the highest point, and the total time taken to hit the ground after leaving the hand is about 4.61 seconds.