A ball of mass 0.4kg, initially at rest, is kicked directly toward a fence from a point 20m away, as shown below. The velocity of the ball as it leaves the kickerś foot is 18m/s at angle of 35 degrees above the horizontal. The top of the fence is 4m high. The ball hits nothing while in flight and air resistence is negligible. The acceleration due to gravity is 9.8 m/s². How far above the top of the fence will the ball pass? Consider the diameter of the ball to be negligible. Answer in units of m. 


To solve this problem, we need to analyze the motion of the ball as a projectile, considering both the horizontal and vertical components of its initial velocity. We can break this down step-by-step:

### Step 1: Resolve the initial velocity into components
Given:
- Initial speed, \( v_0 = 18 \text{ m/s} \)
- Launch angle, \( \theta = 35^\circ \)

The horizontal and vertical components of the initial velocity are:
\[
v_{0x} = v_0 \cos \theta = 18 \cos 35^\circ
\]
\[
v_{0y} = v_0 \sin \theta = 18 \sin 35^\circ
\]

Calculating these values:
\[
v_{0x} = 18 \cos 35^\circ \approx 18 \times 0.819 = 14.74 \text{ m/s}
\]
\[
v_{0y} = 18 \sin 35^\circ \approx 18 \times 0.574 = 10.33 \text{ m/s}
\]

### Step 2: Determine the time to reach the fence
The fence is located 20 m away horizontally. The time \( t \) it takes for the ball to reach the fence can be found using the horizontal motion equation:
\[
x = v_{0x} t
\]
where \( x = 20 \text{ m} \). Solving for \( t \):
\[
t = \frac{x}{v_{0x}} = \frac{20}{14.74} \approx 1.36 \text{ s}
\]

### Step 3: Calculate the vertical position of the ball at time \( t \)
The vertical displacement \( y \) of the ball at time \( t \) is given by:
\[
y = v_{0y} t - \frac{1}{2} g t^2
\]
where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.

Substituting the values:
\[
y = 10.33 \times 1.36 - \frac{1}{2} \times 9.8 \times (1.36)^2
\]
\[
y = 14.05 - 9.07 \approx 4.98 \text{ m}
\]

### Step 4: Determine how far above the fence the ball passes
The top of the fence is 4 m high. The ball's vertical position at the fence is 4.98 m, so the height above the fence is:
\[
\text{Height above the fence} = 4.98 \text{ m} - 4 \text{ m} = 0.98 \text{ m}
\]

Therefore, the ball will pass **0.98 meters above the top of the fence**.

---

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the outcome change if the launch angle were increased to 45 degrees?
2. What would happen if air resistance were considered in this problem?
3. How far would the ball travel horizontally before hitting the ground?
4. What is the maximum height reached by the ball during its flight?
5. How long does the ball stay in the air before hitting the ground?

**Tip:** Breaking projectile motion into horizontal and vertical components simplifies calculations. Use trigonometric functions to find these components accurately.

A ball of mass 0.4kg, initially at rest, is kicked directly toward a fence from a point 20m away. The velocity of the ball as it leaves the kicker's foot is 18m/s at an angle of 35 degrees above the horizontal. The top of the fence is 4m high. How far above the top of the fence will the ball pass? Consider the diameter of the ball to be negligible. Answer in units of m.

Solve the projectile motion problem where a ball of mass 0.4kg is kicked from 20m away with an initial velocity of 18m/s at a 35-degree angle. Learn how to calculate how far above a 4m-high fence the ball will pass using kinematic equations and trigonometric functions. Suitable for students in Grades 9-12.

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