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

YES, the motion of the ball is a free fall because the only force acting on it is gravity, assuming air resistance is negligible.

To calculate the final velocity ($v$) of the ball when it hits the floor, we can apply the equation of motion: $v^2 = u^2 + 2gh$ where:

• $u = 8 \, \text{m/s}$ (initial velocity),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $h = 1.8 \, \text{m}$ (height).

Plugging in the values:

Using the equation of motion: $s = ut + \frac{1}{2}gt^2$ where:

• $s = 1.8 \, \text{m}$ (height),
• $u = 8 \, \text{m/s}$,
• $g = 9.8 \, \text{m/s}^2$.

Rearranging the equation to solve for $t$: $1.8 = 8t + \frac{1}{2}(9.8)t^2.$

This is a quadratic equation in the standard form: $4.9t^2 + 8t - 1.8 = 0.$

Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4.9$, $b = 8$, and $c = -1.8$. Calculating gives: $t \approx 0.21 \, \text{s}.$

The velocity after bouncing is reduced by 20%. Thus, the new velocity ($v'$) is: $v' = v(1 - 0.2) = 9.96 \times 0.8 = 7.968 \, \text{m/s}.$ Using the maximum height formula: $h = \frac{v'^2}{2g} = \frac{(7.968)^2}{2(9.8)} \approx 3.25 \, \text{m}.$ The total height the ball would reach is the height of the bounce plus the original height from which it was thrown: $H = 1.8 + 3.25 = 5.05 \, \text{m}.$ Since the ceiling is at 3.5 m, the ball will reach the ceiling after the first bounce.

The velocity-time graph will have the following features:

• Initial velocity starts at $-8 \, \text{m/s}$ (downwards).
• The graph will slope downwards until it hits the floor at $-9.96 \, \text{m/s}$.
• After the bounce, the velocity will start at $7.968 \, \text{m/s}$ (upwards).
• The graph will slope downwards until it returns to 0 m/s at the maximum height.
• Clearly indicate the points for the initial velocity, when the ball hits the floor, and when it reaches the maximum height.