Let's solve this step by step based on the question provided:

Given:

1. Time for the object to fall: $t = 0.5 \, \text{s}$
2. Horizontal distance traveled: $x = 5 \, \text{m}$
3. Gravitational acceleration: $g = 10 \, \text{m/s}^2$

(a) Height of the throw:

The vertical motion is governed by the equation: $h = \frac{1}{2} g t^2$ Substitute $g = 10 \, \text{m/s}^2$ and $t = 0.5 \, \text{s}$: $h = \frac{1}{2} \cdot 10 \cdot (0.5)^2 = \frac{1}{2} \cdot 10 \cdot 0.25 = 1.25 \, \text{m}$

Answer (a): Height = $1.25 \, \text{m}$

(b) Initial horizontal velocity:

The horizontal motion equation is: $x = v_x \cdot t$ Rearranging for $v_x$: $v_x = \frac{x}{t}$ Substitute $x = 5 \, \text{m}$ and $t = 0.5 \, \text{s}$: $v_x = \frac{5}{0.5} = 10 \, \text{m/s}$

Answer (b): Initial velocity = $10 \, \text{m/s}$

(c) Velocity when the object hits the ground:

The velocity components are:

• Horizontal: $v_x = 10 \, \text{m/s}$
• Vertical (using $v_y = g \cdot t$): $v_y = 10 \cdot 0.5 = 5 \, \text{m/s}$

The total velocity $v$ is: $v = \sqrt{v_x^2 + v_y^2}$ Substitute $v_x = 10$ and $v_y = 5$: $v = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 11.18 \, \text{m/s}$

Answer (c): Velocity = $11.18 \, \text{m/s}$

(d) Velocity after 0.2 seconds:

1. Vertical velocity after $t = 0.2 \, \text{s}$: $v_y = g \cdot t = 10 \cdot 0.2 = 2 \, \text{m/s}$
2. Horizontal velocity remains constant: $v_x = 10 \, \text{m/s}$

The total velocity $v$: $v = \sqrt{v_x^2 + v_y^2}$ Substitute $v_x = 10$ and $v_y = 2$: $v = \sqrt{10^2 + 2^2} = \sqrt{100 + 4} = \sqrt{104} = 10.2 \, \text{m/s}$

Answer (d): Velocity = $10.2 \, \text{m/s}$

(e) Angle between velocity and horizontal after 0.2 seconds:

The angle $\theta$ is given by: $\theta = \tan^{-1} \left( \frac{v_y}{v_x} \right)$ Substitute $v_y = 2 \, \text{m/s}$ and $v_x = 10 \, \text{m/s}$: $\theta = \tan^{-1} \left( \frac{2}{10} \right) = \tan^{-1} (0.2)$ Using a calculator: $\theta \approx 11.31^\circ$

Answer (e): Angle = $11.31^\circ$

Summary of Answers:

(a) $h = 1.25 \, \text{m}$
(b) $v_x = 10 \, \text{m/s}$
(c) $v = 11.18 \, \text{m/s}$
(d) $v = 10.2 \, \text{m/s}$
(e) $\theta = 11.31^\circ$

Would you like further explanations or visualizations for any of these answers?
Here are five questions to deepen your understanding:

1. How would the results change if the object were thrown at an angle instead of horizontally?
2. What would happen if air resistance were included in this problem?
3. How is the vertical motion independent of the horizontal motion in this scenario?
4. Can you derive the equations of motion for this kind of problem?
5. How does the value of gravitational acceleration affect the outcome?

Tip: Always separate the horizontal and vertical motions when solving projectile problems to simplify your calculations!