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To find the vertical component of the plane's take-off velocity, we can use trigonometry. The total velocity of the plane is given as 110 km/h, and the angle of take-off is 34°.

The vertical component \( v_{\text{vertical}} \) of the velocity can be found using the sine function:
\[
v_{\text{vertical}} = v \cdot \sin(\theta)
\]
where:
- \( v \) is the magnitude of the velocity (110 km/h),
- \( \theta \) is the angle of take-off (34°).

Substitute the given values:
\[
v_{\text{vertical}} = 110 \cdot \sin(34^\circ)
\]
Using a calculator:
\[
v_{\text{vertical}} = 110 \cdot 0.5592 \approx 61.51 \, \text{km/h}
\]

So, the vertical component of the plane's take-off velocity is approximately **61.51 km/h**.

Would you like any further clarification or more details?

Here are five related questions:
1. How would you find the horizontal component of the velocity?
2. What role does the vertical component play in the plane's ascent?
3. If the angle of take-off increased, how would that affect the vertical component?
4. How would you convert the velocity from km/h to m/s?
5. What is the significance of using sine versus cosine for different components of velocity?

**Tip:** Always remember to use sine for the vertical component and cosine for the horizontal component when resolving a vector in projectile motion or take-off problems.

A plane takes off at 110 km/h at an angle of 34°. Find the vertical component of its take-off velocity.

Learn how to calculate the vertical component of a plane's take-off velocity using trigonometric functions. Given a plane's velocity of 110 km/h and an angle of 34°, this step-by-step solution uses the formula v_vertical = v * sin(θ) to find the vertical velocity component. Ideal for high school students in Grades 10-12.

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