Define the term projectile. Calculate the: The value of X. Speed V with which the ball hits the ground. Calculate the value of Vy. Draw a position-time graph for ball Q from the moment it was vertically thrown upwards until it strikes the ground. Take the ground as the zero position.

The height $X$ is the magnitude of the displacement.

4.2.2 Speed V with which the ball hits the ground

We will use the kinematic equation for final velocity.
Given:
Initial velocity ($v_i$) = $17.5 \text{ m}\cdot\text{s}^{-1}$
Time ($t$) = $5 \text{ s}$
Acceleration due to gravity ($a$) = $-9.8 \text{ m}\cdot\text{s}^{-2}$

The speed $V$ is the magnitude of the final velocity.

4.3 Calculate the value of $V_Y$.

For ball Q, it is thrown vertically upwards from the same height $X$. Let's take upwards as the positive direction.
Given:
Displacement ($\Delta y$) = $-X = -35 \text{ m}$ (since it ends up $35 \text{ m}$ below its starting point)
Acceleration due to gravity ($a$) = $-9.8 \text{ m}\cdot\text{s}^{-2}$
Final speed = $\frac{6}{7} V = \frac{6}{7} (31.5) = 27 \text{ m}\cdot\text{s}^{-1}$
Since the ball hits the ground, its final velocity ($v_f$) is downwards: $v_f = -27 \text{ m}\cdot\text{s}^{-1}$

4.4 Draw a position-time graph for ball Q from the moment it was thrown upwards until it strikes the ground. Take the ground as the zero position.

For ball Q:
Initial position ($y_i$) = $X = 35 \text{ m}$ (since the ground is the zero position)
Initial velocity ($v_{iQ}$) = $V_Y \approx 6.56 \text{ m}\cdot\text{s}^{-1}$
Acceleration ($a$) = $-9.8 \text{ m}\cdot\text{s}^{-2}$

Key points for the graph:

• Initial point: At $t=0 \text{ s}$, $y = 35 \text{ m}$. So, $(0, 35)$.
• Time to reach maximum height ($t_{peak}$): At maximum height, $v_f = 0$. $$v_f = v_{iQ} + at_{peak}$$ $$0 = 6.56 + (-9.8)t_{peak}$$ $$t_{peak} = \frac{6.56}{9.8} \approx 0.67 \text{ s}$$
• Maximum height ($y_{max}$): $$y_{max} = y_i + v_{iQ}t_{peak} + \frac{1}{2}at_{peak}^2$$ $$y_{max} = 35 + (6.56)(0.67) + \frac{1}{2}(-9.8)(0.67)^2$$ $$y_{max} \approx 35 + 4.3952 - 2.201$$ $$y_{max} \approx 37.19 \text{ m}$$ So, the peak of the trajectory is at approximately $(0.67 \text{ s}, 37.19 \text{ m})$.
• Time to hit the ground ($t_{ground}$): At the ground, $y_f = 0 \text{ m}$. $$\Delta y = y_f - y_i = 0 - 35 = -35 \text{ m}$$ $$\Delta y = v_{iQ}t_{ground} + \frac{1}{2}at_{ground}^2$$ $$-35 = (6.56)t_{ground} + \frac{1}{2}(-9.8)t_{ground}^2$$ $$-35 = 6.56t_{ground} - 4.9t_{ground}^2$$ $$4.9t_{ground}^2 - 6.56t_{ground} - 35 = 0$$ Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $$t_{ground} = \frac{-(-6.56) \pm \sqrt{(-6.56)^2 - 4(4.9)(-35)}}{2(4.9)}$$ $$t_{ground} = \frac{6.56 \pm \sqrt{43.0336 + 686}}{9.8}$$ $$t_{ground} = \frac{6.56 \pm \sqrt{729.0336}}{9.8}$$ $$t_{ground} = \frac{6.56 \pm 27.0006}{9.8}$$ Since time must be positive: $$t_{ground} = \frac{6.56 + 27.0006}{9.8} = \frac{33.5606}{9.8} \approx 3.42 \text{ s}$$ So, the ball hits the ground at approximately $(3.42 \text{ s}, 0 \text{ m})$.

The graph will be a downward-opening parabola starting at $(0, 35)$, reaching a maximum height at approximately $(0.67, 37.19)$, and ending at $(3.42, 0)$.

graph TD
    A[Start (0, 35)] --> B[Peak (0.67, 37.19)]
    B --> C[End (3.42, 0)]

    style A fill:#fff,stroke:#333,stroke-width:2px
    style B fill:#fff,stroke:#333,stroke-width:2px
    style C fill:#fff,stroke:#333,stroke-width:2px

    subgraph Position-Time Graph for Ball Q
        direction LR
        t0[0] -- x-axis (Time in s) --> t1[0.67]
        t1 -- x-axis (Time in s) --> t2[3.42]

        y0[0] -- y-axis (Position in m) --> y1[35]
        y1 -- y-axis (Position in m) --> y2[37.19]

        point0(0, 35)
        pointPeak(0.67, 37.19)
        pointEnd(3.42, 0)

        point0 --- pointPeak
        pointPeak --- pointEnd
    end

The graph would look like a parabola opening downwards, starting at a positive position (35 m), rising slightly to a maximum, and then curving downwards to reach 0 m at approximately 3.42 seconds.