Let's break down and solve the given equations step-by-step:

1. $S_y = \frac{15^2 \times \sin^2(30^\circ)}{2 \times 9.8}$

We have:

• $\sin(30^\circ) = 0.5$
• $g = 9.8 \, \text{m/s}^2$

Now, calculating: $S_y = \frac{15^2 \times (0.5)^2}{2 \times 9.8}$ $S_y = \frac{225 \times 0.25}{19.6} = \frac{56.25}{19.6} \approx 2.87 \, \text{m}$

2. $t_{\text{tertinggi}} = \frac{15 \times \sin(30^\circ)}{9.8}$

Substituting values: $t_{\text{tertinggi}} = \frac{15 \times 0.5}{9.8} = \frac{7.5}{9.8} \approx 0.77 \, \text{seconds}$

3. $S_x = \frac{15^2 \times \sin(60^\circ)}{9.8}$

Here:

• $\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$

Calculating: $S_x = \frac{15^2 \times 0.866}{9.8} = \frac{225 \times 0.866}{9.8} = \frac{194.85}{9.8} \approx 19.88 \, \text{m}$

4. $t_{\text{terjauh}} = \frac{2 \times 15 \times \sin(30^\circ)}{9.8}$

Substituting values: $t_{\text{terjauh}} = \frac{2 \times 15 \times 0.5}{9.8} = \frac{15}{9.8} \approx 1.53 \, \text{seconds}$

These are the solutions for each equation. Do you want further details on any step or more explanation? Here are some follow-up questions to consider:

1. How would the results change if the angle was 45° instead of 30°?
2. What physical scenario do these equations represent?
3. How does varying the initial velocity affect the maximum height and range?
4. What happens to the time of flight if the gravitational constant changes (e.g., on a different planet)?
5. Can we derive these equations from basic kinematic equations?

Tip: Memorizing key trigonometric values for common angles like 30°, 45°, and 60° will help speed up calculations in projectile motion problems.