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 find the maximum height reached by stone A, we will first determine how high it rises from the point of projection. Using the kinematic equation for uniformly accelerated motion:

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

where:

• $v$ = final velocity (0 m/s at the maximum height),
• $u$ = initial velocity (10 m/s),
• $a$ = acceleration (-9.81 m/s², acting downwards),
• $s$ = height above the point of projection.

Setting the final velocity $v = 0$ at the maximum height, we have:

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

This gives:

$s = \frac{(10)^2}{2 \times 9.81} \approx 5.1 m.$

The total height above the ground is:

$H_{total} = height_{building} + height_{rise} = 40 m + 5.1 m = 45.1 m.$

At the maximum height, the acceleration of stone A is due to gravity. Thus, the magnitude is 9.81 m/s², directed downwards.

To find the time $x$ after which stone B is dropped from rest, we can use the formula for the distance fallen by stone B:

$d_B = \frac{1}{2} g t^2 = \frac{1}{2} (9.81) t^2$

For stone A, the height when both stones meet is 29.74 m, so:

1. For stone A (upwards):

   • Time taken to reach this point is given by: $29.74 = 40 + 10t - \frac{1}{2}(9.81)t^2$ Simplifying gives us a quadratic equation to solve for t.

2. For stone B (downwards):

   • Since it's dropped from rest, the height calculation involves beginning from its release:
   • Its motion starts after $x$ seconds, so: $29.74 = \frac{1}{2} g(t - x)^2$

Setting these two equations equal at the point they pass each other will allow us to solve for $x$.