a ball is shot at a ground level at an angle of 60 degrees with an initial velocity of 100m/s

To analyze the motion of the ball, we can break the problem into two components: horizontal and vertical. Here are the relevant equations and steps based on the given dataTo analyze the motion of the ball, we can break the problem into two components: horizontal and vertical. Here are the relevant equations and steps based on the given 

### Given:
- Initial velocity \( v_0 = 100 \, \text{m/s} \)
- Angle of projection \( \theta = 60^\circ \)
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)

### 1. **Horizontal and Vertical Components of Initial Velocity:**
- Horizontal component of the initial velocity:
  \[
  v_{0x} = v_0 \cdot \cos(\theta) = 100 \cdot \cos(60^\circ) = 100 \cdot 0.5 = 50 \, \text{m/s}
  \]

- Vertical component of the initial velocity:
  \[
  v_{0y} = v_0 \cdot \sin(\theta) = 100 \cdot \sin(60^\circ) = 100 \cdot \frac{\sqrt{3}}{2} \approx 86.6 \, \text{m/s}
  \]

### 2. **Time of Flight (T):**
The time of flight is the total time the ball stays in the air. It can be calculated using the vertical motion equation, where the displacement is zero (since it returns to the ground):

\[
T = \frac{2 \cdot v_{0y}}{g} = \frac{2 \cdot 86.6}{9.8} \approx 17.7 \, \text{seconds}
\]

### 3. **Maximum Height (H):**
The maximum height is reached when the vertical velocity becomes zero. Using the vertical motion equation:

\[
H = \frac{v_{0y}^2}{2g} = \frac{(86.6)^2}{2 \cdot 9.8} \approx 382.7 \, \text{m}
\]

### 4. **Range (R):**
The range is the horizontal distance the ball travels, and it can be calculated using the horizontal component of the velocity and the time of flight:

\[
R = v_{0x} \cdot T = 50 \cdot 17.7 = 885 \, \text{m}
\]

### Summary of results:
- Time of flight: \( 17.7 \, \text{seconds} \)
- Maximum height: \( 382.7 \, \text{m} \)
- Range: \( 885 \, \text{m} \)

Would you like more details or explanations on any part?

Here are 5 related questions to deepen your understanding:
1. How does the angle of projection affect the time of flight and range?
2. What happens to the maximum height if the initial velocity increases?
3. How would the results change if the ball was shot from a height above ground level?
4. How would air resistance affect the range and time of flight?
5. Can you derive the general formulas for projectile motion?

**Tip**: In projectile motion, increasing the launch angle often increases the height but may reduce the range, especially beyond 45°.

Explore the solution for a projectile motion problem where a ball is shot at a 60-degree angle with an initial velocity of 100 m/s. The solution covers key concepts like time of flight, range, and maximum height. Suitable for high school students (Grades 9-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation