A small ball is dropped from a height of 2 m and bounces a few times after landing on a cement floor - NSC Physical Sciences - Question 3 - 2020 - Paper 1

From the graph, the ball is in contact with the floor from time 0 seconds to approximately 0.64 seconds. Hence, the time of contact is 0.64 seconds.

The time of the first bounce can be determined from the graph. It appears the ball reaches its maximum height approximately at 1.07 seconds (0.64 s + 0.43 s for ascent) after the first impact.

To calculate the speed as the ball leaves the floor, we can use the equation for free fall:

v = rac{y_f - y_i}{t}

Here, $y_f$ is the maximum height (approx. 1.8 m), $y_i$ is the lowest position (0 m), and $t$ is the time (0.43 seconds). Hence,

v = rac{1.8 ext{ m} - 0 ext{ m}}{0.43 ext{ s}}

Calculating the above gives a speed of approximately 4.19 m/s.

The total time indicated on the graph for the motion of the ball is approximately 2.97 seconds, as noted from the x-axis of the graph.