A ball, of mass 0.06 kg, is thrown vertically upwards from the balcony of a building, 3 m above the ground - NSC Physical Sciences - Question 3 - 2021 - Paper 1

The acceleration of the ball at time t = 1,02 s is equal to the acceleration due to gravity, which is approximately $9.81 \, m/s^2$ acting downwards.

To find the difference between areas A₁ and A₂, subtract the area A₂ from area A₁. The specific values will depend on the measurements from the graph, which should be calculated.

Using the velocity at the moment of launch, we can derive this using kinematics. The initial speed can be determined using the graph, where the velocity at time t = 0 can be noted.

The height h can be calculated using kinematic equations. Assuming no air resistance, the maximum height can be calculated from the initial speed and the acceleration due to gravity. The formula is given as: $h = v_i t + \frac{1}{2} a t^2$ where $v_i$ is the initial velocity, $a$ is acceleration, and $t$ is the time until the ball stops moving upwards.

The work done by the ground on the ball can be calculated using the formula for work done, which is given by: $W = F \cdot d$, where $F$ is the force exerted by the ground and $d$ is the displacement of the ball. Since the ball bounces back after hitting the ground, consider the kinetic energy converted to potential energy as it rises.