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 at which the ball hits the ground, we use the equation of motion:

$v_f^2 = v_i^2 + 2a imes d$

Where:

• $v_f$ = final velocity (speed at impact)
• $v_i$ = initial velocity (0 m/s, since it is dropped)
• $a$ = acceleration (9.81 m/s²)
• $d$ = distance fallen (20 m)

Substituting values:

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

$v_f = ext{sqrt}(392) \\ v_f \\approx 19.8 \, \text{m/s}$

To calculate the time taken to reach the ground, we can use the formula:

$d = v_i imes t + \frac{1}{2} a t^2$

Given that the initial velocity $v_i = 0$, the equation simplifies to:

$20 = \frac{1}{2} (9.81) t^2$

Rearranging gives:

$t^2 = \frac{20 imes 2}{9.81} \\ t^2 = 4.08 \\ t \approx 2.02 \, \text{s}$

The velocity-time graph for the free fall of the ball would be a straight line beginning at the origin and sloping upwards to the right. This indicates that as time increases, the velocity of the ball also increases due to the constant acceleration of gravity. The slope of the line would represent the acceleration effect, which is approximately 9.81 m/s².