A ball is dropped from the top of a building 20 m high - NSC Physical Sciences - Question 3 - 2016 - Paper 1

To find the speed of the ball at the moment it hits the ground, we can use the equation of motion:

$v_f^2 = v_i^2 + 2a s$ where:

• $v_f$ is the final velocity,
• $v_i$ is the initial velocity (0 m/s, as the ball starts from rest),
• $a$ is the acceleration due to gravity (approximately 9.81 m/s²), and
• $s$ is the distance fallen (20 m).

Substituting the values:

$v_f^2 = 0 + 2(9.81)(20)$
$v_f^2 = 392.4$

To calculate the time taken for the ball to reach the ground, we can use the equation:

$s = v_i t + \frac{1}{2} a t^2$
Substituting the known values (and recalling that $v_i = 0$):

$-20 = 0 + \frac{1}{2}(-9.81)t^2$
Solving for t:

$-20 = -4.905t^2$

The velocity-time graph for the ball in free fall would be a straight line starting from the origin (0, 0) and sloping upwards to the right. This indicates that as time progresses, the velocity of the ball increases linearly due to the constant acceleration of gravity. The graph has a positive slope, representing the increase in velocity until it hits the ground.