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 approximately $-9.81 \, ext{m/s}^2$, directed downwards due to gravity.

The difference in areas $A_1$ and $A_2$ represents the change in velocity. If $A_1$ and $A_2$ are calculated from the graph, the difference is the numerical value of the change in velocity.

3.4.1 Speed at which the ball is thrown upwards:

Using the kinematic equations, we can determine the initial velocity $u$ as follows:

Assuming the ball reaches the maximum height, we have:

$v^2 = u^2 + 2a s$

Where:

• $v$ = final velocity (0 m/s at max height)
• $a = -9.81 \, ext{m/s}^2$ (downward acceleration due to gravity)
• $s$ = height ($h$)

3.4.2 Height h:

Using the position function, we can find the height as:

$s = ut + \frac{1}{2} a t^2$

Where:

• t = time until it hits the ground.

The work done by the ground can be calculated using the formula:

$W = F \cdot d$

Where:

• $F$ is the force exerted by the ground
• $d$ is the distance the ball travels upwards after bouncing. Using the maximum height reached after the bounce, we can substitute the values to find the work done.