A ball is thrown vertically upwards, with velocity v, from the edge of a roof of a 40 m tall building - NSC Physical Sciences - Question 3 - 2019 - Paper 1

To find the initial velocity, we can use the kinematic equation:
$v = u + gt$
Given that the time taken to reach maximum height is 1.53 s and the acceleration due to gravity is -9.81 m/s² (downward), we can rearrange the formula:
$u = v - gt$
At maximum height, the final velocity (v) is 0 m/s. Thus:
$u = 0 - (9.81)(1.53)$
Calculating this:
$u = -15.04 ext{ m/s}$
Taking upward direction as positive:
The magnitude of the initial velocity is approximately 15.04 m/s.

Using the kinematic equation for vertical motion:
$h = ut + \frac{1}{2}gt²$
We can substitute the values for u, t, and g:
$h = (15.04)(1.53) + \frac{1}{2}(-9.81)(1.53)²$
Calculating further:
$h = 23.04 - 11.47$
Thus, the maximum height reached by the ball above the roof is approximately 11.57 m.

Using the same kinematic equation:
$y = ut + \frac{1}{2}gt²$
Where:

• $u = 15.04 ext{ m/s}$
• $t = 4 ext{ s}$
• $g = -9.81 ext{ m/s}²$
  Substituting these values:
  $y = (15.04)(4) + \frac{1}{2}(-9.81)(4)²$
  Calculating gives:
  $y = 60.16 - 78.48$
  Thus, the position of the ball after 4 seconds is -18.32 m, which means it is approximately 18.32 m below the edge of the roof.

NO. The calculations for the initial velocity and the maximum height reached are based solely on the initial velocity and time taken to reach maximum height, and these values are not influenced by the height of the building. Therefore, answers to 3.2 and 3.3 remain unchanged.