A frog leaps at 6.32 meters per second at a 36.3 degree angle to the horizon. How long does it take for the frog to reach its maximum height? Time equals blank seconds; fill in the blank.

To find the time it takes for the frog to reach its maximum height, we can focus on the vertical component of the frog's velocity. The vertical velocity component \( v_y \) is determined by the initial velocity and the angle of projection.

### Step-by-Step Calculation:

1. **Given Values:**
   - Initial velocity: \( v = 6.32 \, \text{m/s} \)
   - Angle of projection: \( \theta = 36.3^\circ \)
   - Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \)

2. **Vertical Component of Velocity**:
   The vertical component of the velocity \( v_y \) is given by:
   \[
   v_y = v \cdot \sin(\theta)
   \]
   Substituting the known values:
   \[
   v_y = 6.32 \cdot \sin(36.3^\circ)
   \]

3. **Time to Reach Maximum Height**:
   The time to reach maximum height \( t_{\text{max}} \) is the time it takes for the vertical velocity to become zero, which is calculated using the equation:
   \[
   v_y = g \cdot t_{\text{max}}
   \]
   Rearranging to solve for \( t_{\text{max}} \):
   \[
   t_{\text{max}} = \frac{v_y}{g}
   \]

Let me compute this value for you.The time it takes for the frog to reach its maximum height is approximately \( 0.382 \) seconds.

Would you like a more detailed explanation, or do you have any further questions?

Here are 5 related questions for further exploration:
1. What is the maximum height the frog reaches during the leap?
2. How far horizontally does the frog travel before reaching the maximum height?
3. What is the total time of flight for the frog's leap?
4. What is the horizontal distance covered by the frog during the entire jump?
5. How would the time to maximum height change if the angle of projection was increased?

**Tip:** In projectile motion, the time to reach maximum height is always half of the total flight time.

Learn how to calculate the time it takes for a frog to reach its maximum height during a leap at 6.32 m/s with a 36.3-degree angle to the horizon. This solution covers key concepts in projectile motion, including velocity components and time to maximum height. Suitable for students in Grades 10-12.

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