Stone A is thrown vertically upwards with a speed of 10 m·s⁻¹ from the edge of the roof of a 40 m high building, as shown in the diagram below - NSC Physical Sciences - Question 3 - 2019 - Paper 1

To calculate the maximum height reached by stone A, we can use the following kinematic equation:

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

Where:

• $v$ = final velocity (0 m/s at maximum height).
• $u$ = initial velocity (10 m/s upwards).
• $a$ = acceleration due to gravity (-9.81 m/s²).
• $s$ = height above the starting point.

Rearranging the equation for $s$ gives:

$0 = (10)^2 + 2(-9.81)s$

Solving for $s$ yields:

$s = \frac{(10)^2}{2 \times 9.81} \approx 5.10 \, \text{m}$

Since the height of the building is 40 m, the maximum height above the ground will be:

$\text{Total Height} = 40 + 5.10 \approx 45.10 \, \text{m}$

At maximum height, the magnitude of the acceleration of stone A remains constant at 9.81 m/s² directed downwards, due to the force of gravity acting on it.

Since stone B is dropped from rest, its initial velocity is 0 m/s. The time taken for stone B to fall a height of 40 m can be calculated using:

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

Here, $u=0$, $s=40$, and $a=9.81\text{m/s}^2$.

Thus:

$40 = 0 + \frac{1}{2} (9.81) t^2$

Solving gives:

$t^2 = \frac{80}{9.81} \approx 8.16 \Rightarrow t \approx 2.86 \, \text{s}$

Therefore, stone B reaches the ground in approximately 2.86 s after it is dropped.

Knowing that stone A passes stone B when both stones are 29.74 m above the ground, we can find the time taken for each:

1. For stone A:

   Using the height equation:

   $s = ut + \frac{1}{2} (-9.81) t^2$

   Setting $s = 29.74 - 40 = -10.26$ m gives:

   $-10.26 = (10)t + \frac{1}{2} (-9.81)t^2$

   Rearranging leads to:

   $0 = 4.905t^2 - 10t - 10.26$

   Using the quadratic formula, we can find $t$ for stone A.

2. For stone B dropping after x seconds:

   As calculated earlier, the time to reach 40 m is approximately 2.86 s. Thus:

   $x + t_2 = 2.86$

   Where $t_2$ is the time for stone A to reach the height of 29.74 m.

Next, we solve for both $t$ values using the earlier established equations and substitute to find the numerical value of x.