A stunt man drives a car at a speed of 25 m/s off a 40-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20∘. How far from the base of the cliff does the car land?

To solve this problem, we can break the car's motion into horizontal and vertical components, treating it as a projectile motion problem. Here's a step-by-step breakdown:

### Given:
- Speed of the car: \( v = 25 \, \text{m/s} \)
- Height of the cliff: \( h = 40 \, \text{m} \)
- Angle of the incline: \( \theta = 20^\circ \)
- Acceleration due to gravity: \( g = 9.81 \, \text{m/s}^2 \)

### Step 1: Resolve the initial velocity into horizontal and vertical components

The car's velocity has two components, based on the incline angle:

- **Horizontal component**:
  \[
  v_x = v \cdot \cos(\theta) = 25 \cdot \cos(20^\circ)
  \]
- **Vertical component**:
  \[
  v_y = v \cdot \sin(\theta) = 25 \cdot \sin(20^\circ)
  \]

### Step 2: Calculate the time of flight

The vertical motion is influenced by gravity. To find the time the car is in the air, we use the vertical displacement equation:
\[
y = v_y t + \frac{1}{2} (-g) t^2
\]
where \( y = -40 \, \text{m} \) (since the car falls downward by 40 meters).

This gives the quadratic equation:
\[
-40 = (25 \sin(20^\circ)) t - \frac{1}{2} (9.81) t^2
\]

We'll solve this quadratic equation to find the time \( t \).

### Step 3: Calculate the horizontal distance

Once we have the time of flight \( t \), the horizontal distance can be calculated using:
\[
x = v_x \cdot t
\]
where \( v_x = 25 \cos(20^\circ) \).

I'll now solve this step-by-step and give you the final result.The car lands approximately **90.62 meters** away from the base of the cliff. The time of flight is around **3.86 seconds**.

Would you like any further details or explanations?

Here are 5 related questions to explore further:
1. How would the landing distance change if the car's speed increased to 30 m/s?
2. What would happen if the angle of the incline were 30° instead of 20°?
3. How would the problem change if the height of the cliff were 50 meters?
4. What if there was air resistance acting on the car's motion?
5. How can we use energy conservation principles to solve this problem?

**Tip**: Breaking down velocity into horizontal and vertical components is crucial in projectile motion problems, as each component behaves independently.

A stunt man drives a car at a speed of 25 m/s off a 40-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20°. How far from the base of the cliff does the car land?

Learn how to solve a projectile motion problem where a stunt driver launches a car at 25 m/s off a 40-meter-high cliff with a 20° incline. The solution covers breaking down velocity components, using kinematic equations, and calculating time of flight and horizontal distance. Suitable for students in Grades 11-12.

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