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A ball of mass 0.3 kg, initially at rest, is
kicked directly toward a fence from a point
20 m away, as shown below.
The velocity of the ball as it leaves the
kicker’s foot is 17 m/s at angle of 53 ◦
above
the horizontal. The top of the fence is 7 m
high. The kicker’s foot is in contact with the
ball for 0.03 s. The ball hits nothing while in
flight and air resistance is negligible.
The acceleration due to gravity is 9.8 m/s
2
.How far above the top of fence will the ball
pass?
Answer in units of m.

Let's break down the problem step-by-step to calculate how far above the fence the ball will pass.

### Given:
- Initial velocity \( v_0 = 17 \, \text{m/s} \)
- Launch angle \( \theta = 53^\circ \)
- Distance to the fence \( x = 20 \, \text{m} \)
- Height of the fence \( h = 7 \, \text{m} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

### Steps:

1. **Decompose the initial velocity**:
   - Horizontal component of velocity: 
     \[
     v_{0x} = v_0 \cos(\theta) 
     \]
   - Vertical component of velocity: 
     \[
     v_{0y} = v_0 \sin(\theta)
     \]

2. **Time to reach the fence**:  
   The ball travels 20 meters horizontally, and the horizontal velocity remains constant since there is no air resistance.
   \[
   t_{\text{fence}} = \frac{x}{v_{0x}} = \frac{20}{v_0 \cos(\theta)}
   \]

3. **Vertical displacement when the ball reaches the fence**:  
   The vertical displacement \( y \) at time \( t_{\text{fence}} \) is given by the equation of motion:
   \[
   y = v_{0y} t_{\text{fence}} - \frac{1}{2} g t_{\text{fence}}^2
   \]
   This will give the height of the ball at the location of the fence.

4. **Height above the fence**:  
   Finally, subtract the height of the fence from the vertical position of the ball at the time it reaches the fence to find the distance above the fence.

I'll compute these steps now.The ball will pass approximately \( 0.815 \, \text{m} \) above the top of the fence.

Would you like more details or explanations on any part of this solution?

Here are 5 related questions to explore:
1. How long is the ball in the air before it hits the ground?
2. What is the maximum height the ball reaches during its flight?
3. What would happen if the ball were kicked at a different angle but with the same initial speed?
4. How far would the ball travel horizontally if the fence were not there?
5. How much time does it take for the ball to reach the maximum height?

**Tip:** Understanding the components of velocity (horizontal and vertical) is crucial for solving projectile motion problems.

A ball of mass 0.3 kg, initially at rest, is kicked directly toward a fence from a point 20 m away. The velocity of the ball as it leaves the kicker’s foot is 17 m/s at an angle of 53° above the horizontal. The top of the fence is 7 m high. The ball hits nothing while in flight and air resistance is negligible. How far above the top of the fence will the ball pass?

Solve a projectile motion problem where a ball is kicked at 17 m/s at an angle of 53° toward a fence 20 meters away. Learn how to calculate how far above a 7-meter-high fence the ball will pass. This problem involves kinematic equations and basic physics concepts, suitable for students in Grades 11-12.

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