The solution appears to be for a physics problem involving kinematics, likely about calculating acceleration and maximum height. What was the original question?

here are the solutions to your questions:

16. To determine the acceleration of the body:

    Step 1: Calculate the time interval ($T$) between consecutive dots.
    The frequency of the tick-tock timer is $50 \text{ Hz}$.

    $$T = \frac{1}{\text{frequency}} = \frac{1}{50 \text{ Hz}} = 0.02 \text{ s}$$

    Step 2: Identify the distances between consecutive dots (segments) and convert them to meters.

    $$s_1 = 2 \text{ cm} = 0.02 \text{ m}$$ $$s_2 = 4 \text{ cm} = 0.04 \text{ m}$$ $$s_3 = 6 \text{ cm} = 0.06 \text{ m}$$ $$s_4 = 8 \text{ cm} = 0.08 \text{ m}$$ $$s_5 = 10 \text{ cm} = 0.10 \text{ m}$$

    Step 3: Calculate the acceleration using the difference in consecutive displacements.
    The acceleration ($a$) can be found using the formula $a = \frac{\Delta s}{T^2}$, where $\Delta s$ is the difference between consecutive segment lengths.
    Let's use $s_2 - s_1$:

    $$\Delta s = s_2 - s_1 = 0.04 \text{ m} - 0.02 \text{ m} = 0.02 \text{ m}$$ $$a = \frac{0.02 \text{ m}}{(0.02 \text{ s})^2}$$ $$a = \frac{0.02 \text{ m}}{0.0004 \text{ s}^2}$$ $$a = \boxed{50 \text{ m/s}^2}$$

17. To calculate the maximum height reached by the ball:

    Step 1: Determine the time taken to reach the maximum height.
    The ball returns to its starting point after $4 \text{ s}$. The time to reach the maximum height is half of the total time.

    $$t_{\text{up}} = \frac{4 \text{ s}}{2} = 2 \text{ s}$$

    Step 2: Determine the initial velocity ($u$) of the ball.
    At maximum height, the final velocity ($v$) is $0 \text{ m/s}$. Using the equation of motion $v = u + at$, with $a = -g = -10 \text{ m/s}^2$ (since gravity acts downwards).

    $$0 = u + (-10 \text{ m/s}^2)(2 \text{ s})$$ $$0 = u - 20 \text{ m/s}$$ $$u = 20 \text{ m/s}$$

    Step 3: Calculate the maximum height ($h_{max}$).
    Using the equation of motion $s = ut + \frac{1}{2}at^2$:

    $$h_{max} = (20 \text{ m/s})(2 \text{ s}) + \frac{1}{2}(-10 \text{ m/s}^2)(2 \text{ s})^2$$ $$h_{max} = 40 \text{ m} - \frac{1}{2}(10 \text{ m/s}^2)(4 \text{ s}^2)$$ $$h_{max} = 40 \text{ m} - 20 \text{ m}$$ $$h_{max} = \boxed{20 \text{ m}}$$

18. To draw the displacement-time graphs for A and B: The given velocity-time graph shows that both objects A and B are moving with constant positive velocities. Object A has a higher constant velocity than object B. For constant velocity, the displacement-time graph is a straight line with a constant positive slope. A higher velocity corresponds to a steeper slope.

    $$\text{Displacement (m)}$$ $$$$