Igmore the effects of air resistance. 1. A body is projected vertically upward from the roof of a building at \( 40 \mathrm{~m} \cdot \mathrm{~s}^{-1} \). It reaches the ground with a speed of \( 60 \mathrm{~m} \cdot \mathrm{~s}^{-1} \). Calculate: a) the height of the building. b) the total time of flight.

Sure, let's solve the problem step by step by ignoring air resistance. We'll use the standard equations of motion under constant acceleration due to gravity. ### Given: - **Initial velocity upwards (\( u \))**: \( 40 \, \text{m/s} \) - **Final velocity when reaching the ground (\( v \))**: \( 60 \, \text{m/s} \) downward - **Acceleration due to gravity (\( g \))**: \( 10 \, \text{m/s}^2 \) (approximated for simplicity) **Note:** We'll take upward as the positive direction. Therefore, the final velocity \( v \) will be negative since it's downward. ### Part (a): Height of the Building We can use the **kinematic equation** that relates velocity and displacement: \[ v^2 = u^2 + 2a s \] Where: - \( v = -60 \, \text{m/s} \) (negative because it's downward) - \( u = 40 \, \text{m/s} \) - \( a = -g = -10 \, \text{m/s}^2 \) (acceleration is downward) - \( s = -h \) (displacement is downward, hence negative) Plugging in the values: \[ (-60)^2 = (40)^2 + 2(-10)(-h) \] \[ 3600 = 1600 + 20h \] \[ 3600 - 1600 = 20h \] \[ 2000 = 20h \] \[ h = \frac{2000}{20} = 100 \, \text{meters} \] **Height of the building (\( h \)) is 100 meters.** ### Part (b): Total Time of Flight The total time of flight can be divided into two phases: 1. **Going Up:** Time to reach the maximum height. 2. **Coming Down:** Time to descend from the maximum height to the ground. **1. Time to Reach Maximum Height (\( t_1 \)):** At maximum height, the final velocity is 0. \[ v = u + at \] \[ 0 = 40 - 10t_1 \] \[ t_1 = \frac{40}{10} = 4 \, \text{seconds} \] **2. Time to Descend from Maximum Height to Ground (\( t_2 \)):** The total displacement from the maximum height to the ground is \( h + h_1 \), where \( h_1 \) is the additional height gained. First, calculate \( h_1 \): \[ h_1 = ut_1 + \frac{1}{2} a t_1^2 = 40 \times 4 + \frac{1}{2} \times (-10) \times 4^2 = 160 - 80 = 80 \, \text{meters} \] Total height from the ground: \[ H = h + h_1 = 100 + 80 = 180 \, \text{meters} \] Now, calculate \( t_2 \) using the equation: \[ s = \frac{1}{2} g t_2^2 \] \[ 180 = \frac{1}{2} \times 10 \times t_2^2 \] \[ 180 = 5 t_2^2 \] \[ t_2^2 = \frac{180}{5} = 36 \] \[ t_2 = \sqrt{36} = 6 \, \text{seconds} \] **Total Time of Flight (\( T \))**: \[ T = t_1 + t_2 = 4 + 6 = 10 \, \text{seconds} \] ### **Summary of Answers:** - **a) Height of the building:** \( 100 \, \text{meters} \) - **b) Total time of flight:** \( 10 \, \text{seconds} \)