A ball of mass 0.5 kg is thrown vertically upwards from the top edge of a building which is 15.3 m high - NSC Physical Sciences - Question 3 - 2023 - Paper 1

To find the initial speed of the ball when projected upwards, we can use the following kinematic equation:

$v^2 = u^2 + 2as$

Where:

• v = final velocity (0 m/s at the maximum height)
• u = initial velocity (what we are solving for)
• a = acceleration due to gravity (-9.81 m/s²)
• s = height reached above the starting point (5.89 m)

Rearranging the formula gives: $0 = u^2 + 2(-9.81)(5.89)$

$u^2 = 2(9.81)(5.89)$ $u = ext{sqrt}(2 imes 9.81 imes 5.89)$ This calculates to: $u \\approx 10.82 \, m/s$

The kinetic energy of the ball can be calculated using the formula:

$KE = \frac{1}{2}mv^2$

1. Before collision (initial kinetic energy) when it hits the ground:

   • m = 0.5 kg
   • v = 11.92 m/s (the velocity just before the impact)
   • $KE_{initial} = \frac{1}{2} \cdot 0.5 \cdot (11.92)^2 = 35.42 \, J$

2. After collision (final kinetic energy) when it has reached maximum height:

   • The final speed is 0 m/s (momentarily at the peak height)
   • $KE_{final} = \frac{1}{2} \cdot 0.5 \cdot 0^2 = 0 \, J$

The amount of kinetic energy lost during the collision: $KE_{lost} = KE_{initial} - KE_{final} = 35.42 - 0 = 35.42 \, J$

To calculate the time taken to reach point P:

1. The height of point P relative to the ground is the sum of the building height and the height above it:

   • 15.3 m + 5.89 m = 21.19 m

2. Using the equation of motion: $s = ut + \frac{1}{2}at^2$ Where:

• s = 21.19 m
• u = 11.92 m/s (initial speed upward)
• a = -9.81 m/s² (acceleration due to gravity)

Substituting the values: $21.19 = 11.92t - \frac{1}{2}(9.81)t^2$

This is a quadratic equation in the standard form: $0 = -4.905t^2 + 11.92t - 21.19$

We can solve this using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Calculation leads to:

• The two roots give us the times when the ball reaches that height after it is thrown and the time it falls back. We take the positive root for the upward journey.

Assuming we solve it accurately, we find: $t \approx 3.66 \, s$

K = 11.92 m/s

L = 21.19 m

t₂ - t₁ = 3.66 s