3.2 The string of length 4 m , in the figure below, attached to a ball of mass 0.5 kg undergoing a non-uniform circular motion in the vertical plane breaks when the ball is at the lowest point. If the ball then hits the ground at a horizontal distance 8 m and vertical distance 20 m from the position where the string breaks, calculate: 3.2 .1 the time t the ball travelled from when the string breaks to when it hits the ground, 3.2 .2 the velocity \( \mathrm{V}_{\mathrm{B}} \) of the ball at the moment the string breaks,

When the string breaks, the ball will be in free fall with an initial horizontal velocity. To find the time \( t \) the ball traveled, we can use the vertical displacement equation. The vertical drop is 20 m and using the equation \( d = \frac{1}{2} g t^2 \), where \( g \approx 9.81 \, \text{m/s}^2 \): \[ 20 = \frac{1}{2} \times 9.81 \times t^2 \Rightarrow t^2 = \frac{20 \times 2}{9.81} \Rightarrow t \approx 2.01 \, \text{s} \] For the horizontal distance of 8 m, we can use the horizontal motion equation: \[ \text{Horizontal velocity} = \frac{\text{Distance}}{\text{Time}} = \frac{8}{t} \approx \frac{8}{2.01} \approx 3.98 \, \text{m/s} \] Next, using Pythagorean theorem we find the ball's velocity \( V_B \) at the moment the string breaks, which combines both horizontal and vertical components. The vertical component can be calculated as \( V_{y} = g \cdot t \), leading to an approximate value of the velocity at breakage using: \[ V_{y} \approx 9.81 \times t \approx 9.81 \times 2.01 \approx 19.7 \, \text{m/s} \] Now, we can calculate \( V_B \): \[ V_B = \sqrt{(3.98)^2 + (19.7)^2} \approx \sqrt{15.84 + 388.09} \approx \sqrt{403.93} \approx 20.1 \, \text{m/s} \] So, the ball’s velocity when the string breaks is approximately \( 20.1 \, \text{m/s} \)!