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Let's break down this projectile motion problem:

### **Given Information:**
- Initial height (\( h_0 \)): 20 m
- Initial velocity (\( v_0 \)): 40 m/s
- Launch angle (\( \theta \)): 45°
- Acceleration due to gravity (\( g \)): 9.8 m/s²
- Components of initial velocity:
  \[
  v_{0x} = v_0 \cos(45°) = \frac{40}{\sqrt{2}} \approx 28.28 \, \text{m/s}
  \]
  \[
  v_{0y} = v_0 \sin(45°) = \frac{40}{\sqrt{2}} \approx 28.28 \, \text{m/s}
  \]

### **Step-by-step Solution:**

#### **a) Time of Flight**
The time of flight can be found using the vertical motion equation:
\[
y(t) = h_0 + v_{0y} t - \frac{1}{2} g t^2
\]
Since the projectile lands on the ground, \( y(t) = 0 \):
\[
0 = 20 + 28.28 t - 4.9 t^2
\]
This is a quadratic equation of the form \( at^2 + bt + c = 0 \):
\[
4.9 t^2 - 28.28 t - 20 = 0
\]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
t = \frac{-(-28.28) \pm \sqrt{(-28.28)^2 - 4 \times 4.9 \times (-20)}}{2 \times 4.9}
\]
\[
t = \frac{28.28 \pm \sqrt{799.87}}{9.8}
\]
Calculating the discriminant:
\[
\sqrt{799.87} \approx 28.28
\]
\[
t = \frac{28.28 \pm 28.28}{9.8}
\]
This gives two solutions: \( t_1 = 0 \) (initial time) and \( t_2 = \frac{56.56}{9.8} \approx 5.77 \, \text{s} \).

So, the time of flight is approximately **5.77 seconds**.

#### **b) Horizontal Range**
The range can be calculated using the horizontal velocity and the time of flight:
\[
R = v_{0x} \times t
\]
\[
R = 28.28 \, \text{m/s} \times 5.77 \, \text{s} \approx 163.24 \, \text{m}
\]

#### **c) Maximum Height**
The maximum height occurs when the vertical velocity is zero. The time to reach maximum height (\( t_{max} \)) is given by:
\[
t_{max} = \frac{v_{0y}}{g} = \frac{28.28}{9.8} \approx 2.88 \, \text{s}
\]
Now, plug this into the vertical motion equation to find the maximum height:
\[
y_{max} = h_0 + v_{0y} t_{max} - \frac{1}{2} g t_{max}^2
\]
\[
y_{max} = 20 + 28.28 \times 2.88 - 4.9 \times (2.88)^2
\]
\[
y_{max} = 20 + 81.45 - 40.61 \approx 60.84 \, \text{m}
\]

### **Final Answers:**
1. **Time of flight:** 5.77 seconds
2. **Horizontal range:** 163.24 meters
3. **Maximum height:** 60.84 meters

Would you like more details on any part of the solution or have additional questions?

### **5 Related Questions:**
1. How would the time of flight change if the initial height was different?
2. How does the launch angle affect the horizontal range of a projectile?
3. What would happen to the maximum height if the initial velocity increased?
4. How can we find the impact velocity of the projectile just before it hits the ground?
5. How do air resistance and other forces affect projectile motion?

### **Tip:**  
For projectile motion problems, always break down the initial velocity into horizontal and vertical components. This simplifies the analysis of motion in each direction independently.

En un punto situado a 20 m sobre el suelo lanzamos un cuerpo a 40 m/s formando un ángulo de 45° con la horizontal y cae sobre el suelo. Calcula: a) El tiempo de vuelo; b) El alcance; c) La altura máxima.

Learn how to calculate the time of flight, horizontal range, and maximum height for a projectile launched from a height of 20 m at 40 m/s with a 45° angle. This detailed solution involves kinematic equations and trigonometry. Suitable for high school students in Grades 10-12.

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