A projectile is launched at ground level with an initial

speed of 32 m/s at an angle of 66.7 degrees above the horizontal.

What is the horizontal component of the initial velocity?

The horizontal component of the initial velocity can be found using the following formula:

\[
v_{x} = v_0 \cdot \cos(\theta)
\]

Where:
- \( v_0 = 32 \, \text{m/s} \) is the initial velocity,
- \( \theta = 66.7^\circ \) is the launch angle.

Now, calculating \( v_x \):

\[
v_{x} = 32 \cdot \cos(66.7^\circ)
\]

Let me calculate that for you.The horizontal component of the initial velocity is approximately \( 12.66 \, \text{m/s} \).

Would you like more details or have any questions about this calculation?

Here are 5 related questions to explore further:
1. What is the vertical component of the initial velocity?
2. How long will the projectile be in the air?
3. What is the maximum height reached by the projectile?
4. How far horizontally will the projectile travel before hitting the ground?
5. What are the effects of air resistance on the projectile’s trajectory?

**Tip:** When breaking velocity into components, always use the correct trigonometric functions—cosine for horizontal and sine for vertical components.

A projectile is launched at ground level with an initial speed of 32 m/s at an angle of 66.7 degrees above the horizontal. What is the horizontal component of the initial velocity?

This solution explains how to calculate the horizontal component of the initial velocity for a projectile launched with a speed of 32 m/s at an angle of 66.7 degrees above the horizontal. Learn about using trigonometric functions in projectile motion problems, suitable for students in Grades 9-12.

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