QUESTION 4 Ball \( \mathbf{A} \) is projected vertically upwards at a velocity of \( 16 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) from the ground. Ignors the effects of air resistance. Use the ground as zero reference. 4.1 Calculate the time taken by ball \( \mathbf{A} \) to return to the ground. 4.2 Sketch a velocity-time graph for ball \( \mathbf{A} \). Show the following on the graph: (a) Initial velocity of ball \( \mathbf{A} \) (b) Time taken to reach the highest point of the motion (c) Time taken to return to the ground ONE SECOND after ball \( \mathbf{A} \) is projected upwards, a second ball, B, is thrown vertically downwards at a velocity of \( 9 \mathrm{~m} \cdot \mathrm{~s}^{-1} \) from a balcony 30 m above the ground. Refer to the diagram. 4.3 Calculate how high above the ground ball \( \mathbf{A} \) will be at the instant the two balls pass each other.

To solve the problem step by step, we will break it down into the three parts as outlined in the question. ### 4.1 Calculate the time taken by ball \( \mathbf{A} \) to return to the ground. **Known conditions:** - Initial velocity of ball \( A \) (\( u_A \)) = \( 16 \, \text{m/s} \) - Acceleration due to gravity (\( g \)) = \( -9.81 \, \text{m/s}^2 \) (negative because it acts downwards) - Final velocity when it returns to the ground (\( v_A \)) = \( 0 \, \text{m/s} \) at the highest point. Using the kinematic equation: \[ s = ut + \frac{1}{2} a t^2 \] where: - \( s \) = displacement (which is \( 0 \) when it returns to the ground) - \( u \) = initial velocity - \( a \) = acceleration - \( t \) = time Since the ball returns to the ground, we can set \( s = 0 \): \[ 0 = 16t - \frac{1}{2} \cdot 9.81 \cdot t^2 \] Rearranging gives: \[ 0 = 16t - 4.905t^2 \] Factoring out \( t \): \[ t(16 - 4.905t) = 0 \] This gives us two solutions: 1. \( t = 0 \) (the time of projection) 2. \( 16 - 4.905t = 0 \) Solving for \( t \): \[ 4.905t = 16 \implies t = \frac{16}{4.905} \approx 3.26 \, \text{s} \] ### 4.2 Sketch a velocity-time graph for ball \( \mathbf{A} \). **Key points to include in the graph:** - Initial velocity (\( u_A = 16 \, \text{m/s} \)) at \( t = 0 \). - The time taken to reach the highest point can be calculated using: \[ v = u + at \implies 0 = 16 - 9.81t \implies t = \frac{16}{9.81} \approx 1.63 \, \text{s} \] - The time taken to return to the ground is \( 3.26 \, \text{s} \). The velocity-time graph will be a straight line starting from \( 16 \, \text{m/s} \) at \( t = 0 \), decreasing to \( 0 \, \text{m/s} \) at \( t \approx 1.63 \, \text{s} \), and then continuing to decrease to \( -16 \, \text{m/s} \) at \( t = 3.26 \, \text{s} \). ### 4.3 Calculate how high above the ground ball \( \mathbf{A} \) will be at the instant the two balls pass each other. **Known conditions for ball \( B \):** - Initial velocity of ball \( B \) (\( u_B \)) = \( -9 \, \text{m/s} \) (downward) - Height of the balcony = \( 30 \, \text{m} \) **Position of ball \( A \) at time \( t \):** Using the equation for displacement: \[ s_A = u_A t + \frac{1}{2} a t^2 = 16t - \frac{1}{2} \cdot 9.81 t^2 \] **Position of ball \( B \) at time \( t \):** \[ s_B = 30 - (9t + \frac{1}{2} \cdot 9.81 t^2) \] Setting \( s_A = s_B \) to find the time when they meet: \[ 16t - \frac{1}{2} \cdot 9.81 t^2 = 30 - (9t + \frac{1}{2} \cdot 9.81 t^2) \] This simplifies to: \[ 16t - 9t = 30 \] \[ 7t = 30 \implies t = \frac{30}{7} \approx 4.29 \, \text{s} \] Now, substituting \( t \) back into the equation for \( s_A \) to find the height: \[ s_A = 16 \left(\frac{30}{7}\right) - \frac{1}{2} \cdot 9.81 \left(\frac{30}{7}\right)^2 \] Calculating \( s_A \): 1. Calculate \( 16 \left(\frac{30}{7}\right) \) 2. Calculate \( \frac{1}{2} \cdot 9.81 \left(\frac{30}{7}\right)^2 \) Let's perform these calculations. ### Performing the calculations 1. \( 16 \left(\frac{30}{7}\right) = \frac{480}{7} \approx 68.57 \) 2. \( \frac{1}{2} \cdot 9.81 \left(\frac{30}{7}\right)^2 = \frac{1}{2} \cdot 9.81 \cdot \frac{900}{49} = \frac{4410}{49} \approx 90.00 \) Now substituting back: \[ s_A = 68.57 - 90.00 \approx -21.43 \, \text{m} \] Since this is negative, it indicates that the balls do not meet before ball \( A \) returns to the ground. ### Conclusion - The time taken by ball \( A \) to return to the ground is approximately \( 3.26 \, \text{s} \). - The velocity-time graph shows a linear decrease from \( 16 \, \text{m/s} \) to \( -16 \, \text{m/s} \). - The height above the ground when the two balls pass each other cannot be calculated as they do not meet before ball \( A \) returns to the ground. If you need further clarification or additional calculations, please let me know!