To determine the vertical velocity at the moment of projection when given an angle of [tex]\(30^\circ\)[/tex] and a horizontal velocity of 50 km/h, follow these steps:
1. Understand the components of projectile motion:
- The horizontal velocity component ([tex]\(V_x\)[/tex]) is given as 50 km/h.
- The vertical velocity component ([tex]\(V_y\)[/tex]) needs to be calculated.
2. Relation between angle and velocity components:
- For projectile motion, the vertical component of the velocity can be found using trigonometric functions. Specifically, the tangent of the angle of projection gives the ratio of the vertical component to the horizontal component:
[tex]\[
\tan(\theta) = \frac{V_y}{V_x}
\][/tex]
- Here, [tex]\(\theta = 30^\circ\)[/tex].
3. Calculate the vertical velocity:
- Using the tangent function for the given angle:
[tex]\[
\tan(30^\circ) = \frac{1}{\sqrt{3}}
\][/tex]
- Hence,
[tex]\[
\frac{V_y}{50 \, \text{km/h}} = \frac{1}{\sqrt{3}}
\][/tex]
4. Solve for [tex]\(V_y\)[/tex]:
- Multiplying both sides by 50 km/h:
[tex]\[
V_y = 50 \, \text{km/h} \times \frac{1}{\sqrt{3}} = 50 \times \frac{1}{\sqrt{3}} \, \text{km/h} = \frac{50}{\sqrt{3}} \, \text{km/h}
\][/tex]
- Simplifying further using exact trigonometric values:
[tex]\[
V_y \approx 28.867513459481287 \, \text{km/h}
\][/tex]
So, the vertical velocity at the moment of projection is approximately [tex]\(28.867513459481287\)[/tex] km/h. Therefore, the most accurate option from the given choices is:
(4) 30 kmph
