A girl stands on a platform in a classroom - NSC Physical Sciences - Question 2 - 2016 - Paper 1

YES. The motion of the ball while moving downwards towards the floor can be classified as free fall because the only force acting on it is gravity after it leaves her hand. There is no other force (such as air resistance) considered in this scenario.

To calculate the velocity of the ball just before it hits the floor, we can use the following kinematic equation:

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

Where:

• $v$ = final velocity
• $u = -8 ext{ m·s}^{-1}$ (initial velocity downward)
• $a = 9.8 ext{ m·s}^{-2}$ (acceleration due to gravity)
• $s = 1.8 ext{ m}$ (distance to the floor)

Substituting the values:

$v^2 = (-8)^2 + 2 imes 9.8 imes 1.8$ $v^2 = 64 + 35.28$ $v^2 = 99.28$ $v = ext{sqrt}(99.28)$ $v \\approx 9.96 ext{ m·s}^{-1}$

Thus, the magnitude of the velocity with which the ball hits the floor is approximately $9.96 ext{ m·s}^{-1}$.

To calculate the time taken for the ball to hit the floor, we can use the kinematic equation:

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

Where:

• $s = 1.8 ext{ m}$
• $u = -8 ext{ m·s}^{-1}$
• $a = 9.8 ext{ m·s}^{-2}$

Substituting the values gives:

$1.8 = -8t + \frac{1}{2} imes 9.8 t^2$

Rearranging results in:

$0 = \frac{1}{2} imes 9.8 t^2 - 8t - 1.8$ $0 = 4.9 t^2 - 8t - 1.8$

Using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$t = \frac{8 \pm \sqrt{(-8)^2 - 4 imes 4.9 imes (-1.8)}}{2 \times 4.9}$

Calculating gives: $t = \frac{8 \pm \sqrt{64 + 35.28}}{9.8}$ $t = \frac{8 \pm \sqrt{99.28}}{9.8}$

Taking the positive root and calculating gives: $t \approx 0.81 ext{ seconds}$.

After the first bounce, the ball's velocity decreases by 20%. Thus, the new upwards velocity is: $v = 9.96 imes (1 - 0.20) = 9.96 imes 0.80 = 7.97 ext{ m·s}^{-1}$

Now, we calculate the maximum height it will reach using the formula: $v^2 = u^2 + 2as$

• Final velocity ($v$) at the maximum height is 0,
• Initial velocity ($u$) is 7.97 m·s⁻¹,
• Acceleration ($a$) is -9.8 m·s² (upwards).

$0 = (7.97)^2 + 2(-9.8)s$ Solving for $s$ gives: $s = \frac{(7.97)^2}{2 imes 9.8} = \frac{63.36}{19.6} \approx 3.24 ext{ m}$

Since the ball will reach a height of 3.24 m upwards and the ceiling is 3.5 m, it will NOT reach the ceiling.

The velocity-time graph will consist of several key points:

1. The initial velocity is -8 m·s⁻¹ (at time t = 0).
2. The velocity reaches -9.96 m·s⁻¹ (when the ball hits the floor at t = 0.81 s).
3. After the bounce, the ball's velocity will change to +7.97 m·s⁻¹ at time t = t_bounce.
4. The ball will decelerate to 0 m·s⁻¹ (at maximum height).
5. The graph will have a downward slope until it hits the floor, a steep upward slope after the bounce, and then a downward slope to the maximum height.