a ball is shot a ground level of an angle of 60 degrees with an initial velocity of 100m/s. a. find the velocity of  ball after 2 seconds b. maximum height of the ball c. range d. time of flight

This problem involves projectile motion. Let's break it down and solve each part step by step.

Given:
- Initial velocity (\( v_0 \)) = 100 m/s
- Angle of projection (\( \theta \)) = 60 degrees
- Gravitational acceleration (\( g \)) = 9.8 m/s²

### Part a: Find the velocity of the ball after 2 seconds

To calculate the velocity after 2 seconds, we need to find both the horizontal and vertical components of the velocity.

1. **Horizontal velocity** (\( v_x \)):
   \[
   v_x = v_0 \cos(\theta) = 100 \times \cos(60^\circ) = 100 \times 0.5 = 50 \, \text{m/s}
   \]
   Horizontal velocity remains constant since there is no horizontal acceleration (ignoring air resistance).

2. **Vertical velocity** (\( v_y \)) after time \( t \):
   \[
   v_y = v_0 \sin(\theta) - g t
   \]
   \[
   v_y = 100 \times \sin(60^\circ) - 9.8 \times 2 = 100 \times 0.866 - 19.6 = 86.6 - 19.6 = 67 \, \text{m/s}
   \]

The total velocity \( v \) is found using the Pythagorean theorem:
\[
v = \sqrt{v_x^2 + v_y^2}
\]
\[
v = \sqrt{50^2 + 67^2} = \sqrt{2500 + 4489} = \sqrt{6989} \approx 83.6 \, \text{m/s}
\]

So, the velocity after 2 seconds is approximately **83.6 m/s**.

### Part b: Maximum height of the ball

The maximum height occurs when the vertical velocity becomes zero. Using the vertical motion formula:
\[
v_y = v_0 \sin(\theta) - g t
\]
At the maximum height, \( v_y = 0 \), so:
\[
0 = v_0 \sin(\theta) - g t
\]
\[
t = \frac{v_0 \sin(\theta)}{g} = \frac{100 \times 0.866}{9.8} = \frac{86.6}{9.8} \approx 8.84 \, \text{seconds} (time to reach maximum height)
\]

Now, using the vertical displacement formula to find the maximum height:
\[
h_{\text{max}} = v_0 \sin(\theta) \times t - \frac{1}{2} g t^2
\]
\[
h_{\text{max}} = 100 \times 0.866 \times 8.84 - \frac{1}{2} \times 9.8 \times (8.84)^2
\]
\[
h_{\text{max}} \approx 766.024 - 381.07 = 384.95 \, \text{meters}
\]

So, the maximum height is approximately **385 meters**.

### Part c: Range of the ball

The range of a projectile is given by:
\[
R = \frac{v_0^2 \sin(2\theta)}{g}
\]
\[
R = \frac{100^2 \sin(120^\circ)}{9.8} = \frac{10000 \times 0.866}{9.8} = \frac{8660}{9.8} \approx 883.7 \, \text{meters}
\]

So, the range is approximately **883.7 meters**.

### Part d: Time of flight

The time of flight is twice the time taken to reach the maximum height:
\[
t_{\text{flight}} = 2 \times t_{\text{max}} = 2 \times 8.84 = 17.68 \, \text{seconds}
\]

So, the total time of flight is approximately **17.68 seconds**.

---

### Summary:
- **Velocity after 2 seconds**: 83.6 m/s
- **Maximum height**: 385 meters
- **Range**: 883.7 meters
- **Time of flight**: 17.68 seconds

Would you like more details on any part? Here are five related questions to extend this topic:

1. How would air resistance affect the range and maximum height?
2. Can you calculate the time when the ball reaches half of the maximum height?
3. What would the velocity be after 5 seconds?
4. How would the results change if the angle of projection were 45 degrees?
5. What is the impact of gravity variation on the range?

**Tip:** Remember, the horizontal velocity component remains constant in the absence of air resistance, while the vertical component is affected by gravity.

A ball is shot at ground level at an angle of 60 degrees with an initial velocity of 100 m/s. a. Find the velocity of the ball after 2 seconds b. Maximum height of the ball c. Range d. Time of flight.

Explore how to solve a projectile motion problem where a ball is launched at 60 degrees with an initial velocity of 100 m/s. Learn to find the velocity after 2 seconds, maximum height, range, and time of flight, using kinematic equations and trigonometry. Suitable for Grades 10-12 students.

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