A boll is lauched uptards into the air at \( 20 \mathrm{~m} / \mathrm{s} \) from the top of a building that is 50 m high . It then taalis past the building and hits the ground. 1. Determine how long it takes for the ball to reach its highest point aboue the ground a. What is the bair's velocity when it passes point? 3. Whion is the same height at which it was lavened from? 4 Detamine the velocity of the ball when it hits the ground? 5. How long does it take to hit the ground 6. What is the balles maximum height from the ground?

Sure, let's analyze the motion of the ball step by step based on the provided information. Here's a summary of the given data and the questions we'll address: **Given:** - **Initial velocity (\( u \))**: \( 20 \ \text{m/s} \) (upwards) - **Initial height (\( h_0 \))**: \( 50 \ \text{m} \) - **Acceleration due to gravity (\( g \))**: \( 9.8 \ \text{m/s}^2 \) (downwards) We'll adopt the convention that **upwards** is the **positive** direction. Let's tackle each question one by one. --- ### **1. Determine how long it takes for the ball to reach its highest point above the ground.** **Solution:** At the highest point, the vertical velocity (\( v \)) of the ball becomes **0 m/s**. Using the kinematic equation: \[ v = u + a t \] Where: - \( v = 0 \ \text{m/s} \) (velocity at the highest point) - \( u = 20 \ \text{m/s} \) - \( a = -g = -9.8 \ \text{m/s}^2 \) (since gravity acts downward) Plugging in the values: \[ 0 = 20 - 9.8 t \\ 9.8 t = 20 \\ t = \frac{20}{9.8} \\ t \approx 2.04 \ \text{seconds} \] **Answer:** It takes approximately **2.04 seconds** for the ball to reach its highest point above the ground. --- ### **2. What is the ball's velocity when it passes the point (assuming "point" refers to the initial launch point at 50 m height) on its way down?** **Solution:** Assuming "point" refers to the **launch height** of \( 50 \ \text{m} \), we need to find the velocity of the ball as it passes this height on its way down. We'll use the principle of **symmetry in projectile motion**, which states that the speed of the ball when passing the same height on the way down is equal in magnitude to its speed when passing that height on the way up. However, since the ball was launched **upwards**, it spends different times ascending and descending due to the initial upward velocity. But to accurately determine the velocity, we'll use the following approach. **Method 1: Conservation of Energy** Alternatively, we can calculate the velocity using energy conservation, but it's simpler to use kinematics here. **Method 2: Kinematic Equations** Let's determine the time it takes to return to the launch height and then find the velocity at that time. 1. **First**, calculate the total time to reach the highest point: \( t_1 = 2.04 \ \text{s} \) 2. **Next**, the time to descend back to the launch height is the same: \( t_2 = t_1 = 2.04 \ \text{s} \) 3. **Total time to reach the launch height again:** \( t = t_1 + t_2 = 4.08 \ \text{s} \) 4. **Now, use the velocity-time relation:** \[ v = u + a t \] \[ v = 20 + (-9.8)(4.08) \] \[ v = 20 - 40 \] \[ v = -20 \ \text{m/s} \] **Note:** The negative sign indicates the direction is downward. **Answer:** When passing the launch height on its way down, the ball's velocity is **20 m/s downward**. --- ### **3. When is the same height at which it was launched from?** **Interpretation:** This seems to ask **at what time** the ball returns to the launch height of **50 m**. **Solution:** As calculated in the previous step, the ball takes: - **2.04 seconds** to reach the highest point. - **2.04 seconds** to descend back to the launch height. **Total time to return to launch height:** \[ t = 2.04 + 2.04 = 4.08 \ \text{seconds} \] **Answer:** The ball returns to the launch height of **50 meters** after approximately **4.08 seconds**. --- ### **4. Determine the velocity of the ball when it hits the ground.** **Solution:** To find the velocity when the ball hits the ground, we'll consider the entire motion from the launch point to the ground. **Approach:** 1. **Determine the total displacement (\( s \)) from the launch point to the ground:** - Since the ball is launched from a height of \( 50 \ \text{m} \) and hits the ground, the total displacement is \( -50 \ \text{m} \) (downwards). 2. **Use the kinematic equation:** \[ v^2 = u^2 + 2 a s \] Where: - \( u = 20 \ \text{m/s} \) (upwards) - \( a = -9.8 \ \text{m/s}^2 \) - \( s = -50 \ \text{m} \) Plugging in the values: \[ v^2 = (20)^2 + 2(-9.8)(-50) \\ v^2 = 400 + 980 \\ v^2 = 1380 \\ v = \sqrt{1380} \\ v \approx 37.15 \ \text{m/s} \] **Direction:** Since the ball is moving downward when it hits the ground, the velocity is **downward**. **Answer:** When the ball hits the ground, its velocity is approximately **37.15 m/s downward**. --- ### **5. How long does it take to hit the ground?** **Solution:** To find the total time taken for the ball to hit the ground, we'll consider its motion from the launch point to the peak and then to the ground. **Approach:** Use the kinematic equation: \[ s = u t + \frac{1}{2} a t^2 \] Where: - \( s = -50 \ \text{m} \) (since the ball moves down from the launch point) - \( u = 20 \ \text{m/s} \) - \( a = -9.8 \ \text{m/s}^2 \) Plugging in the values: \[ -50 = 20 t - 4.9 t^2 \\ 4.9 t^2 - 20 t - 50 = 0 \] **Solving the quadratic equation for \( t \):** \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where: - \( a = 4.9 \) - \( b = -20 \) - \( c = -50 \) \[ t = \frac{20 \pm \sqrt{(-20)^2 - 4(4.9)(-50)}}{2(4.9)} \\ t = \frac{20 \pm \sqrt{400 + 980}}{9.8} \\ t = \frac{20 \pm \sqrt{1380}}{9.8} \\ t = \frac{20 \pm 37.15}{9.8} \] We discard the negative root since time cannot be negative: \[ t = \frac{20 + 37.15}{9.8} \\ t = \frac{57.15}{9.8} \\ t \approx 5.83 \ \text{seconds} \] **Answer:** It takes approximately **5.83 seconds** for the ball to hit the ground. --- ### **6. What is the ball's maximum height from the ground?** **Solution:** The maximum height is achieved at the topmost point of the ball's trajectory. **Approach:** 1. **Calculate the additional height gained while moving upwards:** \[ h_{\text{additional}} = u t + \frac{1}{2} a t^2 \] Where at the highest point: - \( u = 20 \ \text{m/s} \) - \( a = -9.8 \ \text{m/s}^2 \) - \( t = 2.04 \ \text{seconds} \) (time to reach the highest point) \[ h_{\text{additional}} = 20(2.04) + \frac{1}{2}(-9.8)(2.04)^2 \\ h_{\text{additional}} = 40.8 - 0.5 \times 9.8 \times 4.1616 \\ h_{\text{additional}} = 40.8 - 20.416 \\ h_{\text{additional}} = 20.384 \ \text{m} \] 2. **Total maximum height from the ground:** \[ h_{\text{max}} = h_0 + h_{\text{additional}} \] \[ h_{\text{max}} = 50 + 20.384 = 70.384 \ \text{m} \] **Answer:** The ball reaches a **maximum height of approximately 70.38 meters** above the ground. --- ### **Summary of Answers:** 1. **Time to highest point:** ≈ **2.04 seconds** 2. **Velocity when passing the launch height downward:** **20 m/s downward** 3. **Time to return to launch height:** ≈ **4.08 seconds** 4. **Velocity when hitting the ground:** ≈ **37.15 m/s downward** 5. **Total time to hit the ground:** ≈ **5.83 seconds** 6. **Maximum height above ground:** ≈ **70.38 meters**