Stone A is projected vertically upwards at a speed of 12 m.s⁻¹ from a height h above the ground. Ignore the effects of air resistance. Calculate the time taken for stone A to reach its maximum height.

Let upwards be the positive direction.
Initial velocity of stone A, $v_{iA} = +12 \text{ m/s}$.
Final velocity of stone A at maximum height, $v_{fA} = 0 \text{ m/s}$.
Acceleration due to gravity, $a = -9.8 \text{ m/s}^2$.
Using the kinematic equation $v_f = v_i + at$:

$$0 = 12 + (-9.8)t_{maxA}$$ $$9.8 t_{maxA} = 12$$ $$t_{maxA} = \frac{12}{9.8}$$ $$t_{maxA} \approx 1.22\text{ s}$$

The time taken for stone A to reach its maximum height is $\boxed{\text{}1.22\text{ s}\text{}}$.

When stone A reaches its maximum height, the time elapsed is $t_{maxA} = \frac{12}{9.8}\text{ s}$.
At this instant, the speed of stone B is $3v$.
For stone B, let downwards be the positive direction for its initial velocity, or continue with upwards as positive and use negative signs. Let's stick to upwards as positive.
Initial velocity of stone B, $v_{iB} = -v$.
Final velocity of stone B at time $t_{maxA}$, $v_{fB} = -3v$.
Acceleration due to gravity, $a = -9.8 \text{ m/s}^2$.
Using the kinematic equation $v_f = v_i + at$:

$$-3v = -v + (-9.8)\left(\frac{12}{9.8}\right)$$ $$-3v = -v - 12$$ $$-2v = -12$$ $$v = 6\text{ m/s}$$

The speed $v$ with which stone B is thrown downwards is $\boxed{\text{}6\text{ m/s}\text{}}$.

Stone A passes its initial position on its way down at a time $t_{returnA} = 2 \times t_{maxA}$.

$$t_{returnA} = 2 \times \frac{12}{9.8} = \frac{24}{9.8}\text{ s}$$

At this instant, stone B hits the ground.
For stone B:
Initial velocity, $v_{iB} = -v = -6 \text{ m/s}$ (since $v$ is a speed, and it's thrown downwards).
Time, $t = t_{returnA} = \frac{24}{9.8}\text{ s}$.
Acceleration, $a = -9.8 \text{ m/s}^2$.
Displacement, $\Delta y_B = -h$.
Using the kinematic equation $\Delta y = v_i t + \frac{1}{2}at^2$:

$$-h = (-6)\left(\frac{24}{9.8}\right) + \frac{1}{2}(-9.8)\left(\frac{24}{9.8}\right)^2$$ $$-h = -\frac{144}{9.8} - 4.9 \left(\frac{576}{9.8^2}\right)$$ $$-h = -\frac{144}{9.8} - \frac{576}{2 \times 9.8}$$ $$-h = -\frac{144}{9.8} - \frac{288}{9.8}$$ $$-h = -\frac{432}{9.8}$$ $$h = \frac{432}{9.8}$$ $$h \approx 44.08\text{ m}$$

The height $h$ is $\boxed{\text{}44.08\text{ m}\text{}}$.

Both graphs will be straight lines with a constant slope of $-9.8 \text{ m/s}^2$ (acceleration due to gravity).

• For Stone A:

  • Initial velocity: $v_{iA} = 12 \text{ m/s}$.
  • Time to reach maximum height: $t_{maxA} \approx 1.22 \text{ s}$. At this time, velocity is $0 \text{ m/s}$.
  • Time to return to initial height: $t_{returnA} \approx 2.45 \text{ s}$. At this time, velocity is $-12 \text{ m/s}$.
  • Total time to hit the ground (displacement $\Delta y_A = -h = -44.08 \text{ m}$): $$v_{fA}^2 = v_{iA}^2 + 2a\Delta y_A = (12)^2 + 2(-9.8)(-44.08) \approx 1007.97$$ $$v_{fA} \approx -31.75 \text{ m/s}$$ $$t_{totalA} = \frac{v_{fA} - v_{iA}}{a} = \frac{-31.75 - 12}{-9.8} \approx 4.46 \text{ s}$$
  • The graph for stone A starts at $(0, 12)$, passes through $(1.22, 0)$, $(2.45, -12)$, and ends at approximately $(4.46, -31.75)$.

• For Stone B:

  • Initial velocity: $v_{iB} = -v = -6 \text{ m/s}$.
  • Time to hit the ground (when A returns to initial position): $t_{returnA} \approx 2.45 \text{ s}$.
  • Final velocity when hitting the ground: $$v_{fB} = v_{iB} + at_{returnA} = -6 + (-9.8)\left(\frac{24}{9.8}\right) = -6 - 24 = -30 \text{ m/s}$$
  • The graph for stone B starts at $(0, -6)$ and ends at $(2.45, -30)$.

Sketch of Velocity-Time Graphs:

A velocity-time graph would show:

• X-axis: Time (s)
• Y-axis: Velocity (m/s)
• Graph for Stone A (Line A): A straight line starting at $(0, 12)$, sloping downwards, crossing the time axis at $t = \text{}1.22\text{ s}\text{}$ (time to reach maximum height), continuing to $t \approx 2.45 \text{ s}$ where its velocity is $-12 \text{ m/s}$, and ending at approximately $(4.46, -31.75)$.
• Graph for Stone B (Line B): A straight line starting at $(0, \text{} -6\text{ m/s}\text{})$, sloping downwards parallel to Line A, and ending at approximately $(2.45, -30)$.
• Labels:
  • Mark $1.22 \text{ s}$ on the time axis, indicating $t_{maxA}$.
  • Mark $-6 \text{ m/s}$ on the velocity axis, indicating the initial velocity of stone B.
  • Label the lines clearly as "Stone A" and "Stone B".