Ball A is dropped from rest from the top of a building 15.2 m high - NSC Physical Sciences - Question 3 - 2023 - Paper 1

To find the time taken for ball A to strike the ground, we use the kinematic equation:

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

Where:

• $s$ = distance fallen (15.2 m - 3.2 m = 12 m)
• $u$ = initial velocity (0 m/s, as it is dropped)
• $a$ = acceleration due to gravity (approximately $9.81 \, m/s^2$)

Plugging in the values:

$12 = 0 \cdot t + \frac{1}{2}(9.81)t^2$

Calculating:

$12 = 4.905t^2$

Dividing both sides by 4.905:

$t^2 = \frac{12}{4.905} \approx 2.44$

Taking the square root gives:

$t \approx 1.56 \text{ seconds}$

To find the magnitude of the velocity of ball B when it is projected upwards, we need to consider that it reaches the ground at the same time as ball A. Since ball A takes approximately 1.56 seconds to reach the ground:

Using the kinematic equation:

$v = u + at$

Where:

• $v$ = final velocity (which we will find)
• $u$ = initial velocity (unknown initially, we will solve for it)
• $a$ = acceleration due to gravity (acting downward, thus -9.81 m/s²)
• $t$ = time taken (1.56 seconds)

Assuming ball B reaches a height equal to ball A's drop and then falls for the same time:

We set $v = 0$ at the maximum height, solving for $u$:

$0 = u - 9.81(1.56)$

Thus:

$u = 9.81(1.56) ext{ m/s} \\ = 15.33 ext{ m/s}$

1. Graph for Ball A:

   • Starting time: $t = 0$ at 12 m (height after falling 3.2 m).
   • Initial position: 12 m (as it falls from 15.2 m to the ground).
   • Time when ball A strikes the ground: $t = 1.56$ seconds.

2. Graph for Ball B:

   • Starting time: same as Ball A.
   • Initial position: 0 m (ground level).
   • Time when both balls strike the ground is the same (at $t = 1.56$ seconds).