a) the height of the building. b) the total time of flight. A girl stands on a bridge \( 11,25 \mathrm{~m} \) above a boy on the ground. The boy throws an orange vertically upward at \( 10 \mathrm{~m} \mathrm{~s}^{-1} \) and at the same instant the girl drops an apple. Calculate: a) the maximum height obtained by the orange and state whether or not it beaches the girl. b) the vertical distance between the orange and the apple 1 s after they were in motion.

To solve the problem, let's break it down step-by-step. For the orange thrown by the boy, we'll calculate the maximum height it reaches. The initial velocity (\( u \)) is \( 10 \, \text{m/s} \), and the acceleration due to gravity (\( g \)) is approximately \( -9.81 \, \text{m/s}^2 \). 1. Maximum Height of the Orange: Using the formula \( v^2 = u^2 + 2as \) where \( v = 0 \) (at the peak height), we have: \[ 0 = (10)^2 + 2(-9.81)s \] \[ 100 = 19.62s \implies s \approx 5.1 \, \text{m} \] Thus, the maximum height above the boy's hand is \( 5.1 \, \text{m} \). To find the total height above the ground: \[ \text{Total height} = \text{Height of the bridge} + \text{Height of the orange} = 11.25 + 5.1 = 16.35 \, \text{m}. \] Since the maximum height \( (16.35 \, \text{m}) \) is above the bridge’s height, the orange will definitely reach the girl. 2. Total Time of Flight for the Orange: To find the total time until the orange hits the ground, we can use the equation of motion: \[ s = ut + \frac{1}{2} a t^2 \] Here, the total height fallen (from its maximum height) is \( 16.35 \, \text{m} \), \( u = 0 \, \text{m/s} \) from the peak, and \( a = 9.81 \, \text{m/s}^2 \). However, we can also determine the time to the peak: \[ t_{up} = \frac{10}{9.81} \approx 1.02 \, \text{s} \] Then the time of falling back down can be calculated by using total distance fallen: Using \( s = \frac{1}{2} g t^2 \), where \( s = 16.35 \), \[ 16.35 = 4.905 t^2 \implies t^2 = \frac{16.35}{4.905} \implies t \approx 1.80 \, \text{s}. \] Therefore, the total time of flight will be approximately \( 1.02 \, \text{s} + 1.80 \, \text{s} = 2.82 \, \text{s}. \) Now for the Apple dropped: 3. Vertical Distance Between the Orange and Apple after 1 second: The position of the apple after 1 second (from rest) is given by: \[ s = \frac{1}{2} g t^2 \implies \text{(downward)} = \frac{1}{2} \times 9.81 \times (1)^2 \approx 4.905 \text{m}. \] So, the apple is \( 11.25 - 4.905 = 6.345 \, \text{m} \) above ground level. For the orange after 1 second can be calculated as follows: 1. **Upward Movement of the orange:** \[ s = ut - \frac{1}{2} g t^2 = 10 \cdot 1 - 0.5 \cdot 9.81 \cdot 1^2 \approx 10 - 4.905 = 5.095 \text{m} \] So, the orange's height above the ground after 1 second is approximately: \[ 11.25 + 5.095 = 16.345 \, \text{m} \] 4. Now, let's determine the vertical distance between the orange and the apple after 1 second: \[ \text{Distance} = \text{Height of Orange} - \text{Height of Apple} = 16.345 - 6.345 = 10.00 \, \text{m}. \] In conclusion: a) The maximum height of the orange is approximately \( 16.35 \, \text{m} \) and it does reach the girl. The total flight time is \( 2.82 \, \text{s} \). b) The vertical distance between the orange and the apple after \( 1 \, \text{s} \) is \( 10 \, \text{m} \).