A hiker throws a stone from the upper edge of a vertical cliff. The stone's initial velocity is 25.0 m/s directed at 40.0° with the face of the cliff, as shown in the figure. The stone hits the ground 3.75 s after being thrown and feels no appreciable air resistance as it falls. What is the height of the cliff?

To solve this problem, we need to break down the stone's motion into horizontal and vertical components and use kinematic equations to find the height of the cliff.

### Step 1: Breakdown of the Initial Velocity
The initial velocity (\(v_0\)) is 25.0 m/s, directed at an angle of 40.0° with the face of the cliff.

1. **Horizontal component** of the velocity:
   \[
   v_{0x} = v_0 \cdot \cos(40.0^\circ)
   \]
   
2. **Vertical component** of the velocity:
   \[
   v_{0y} = v_0 \cdot \sin(40.0^\circ)
   \]

### Step 2: Vertical Motion Analysis
The vertical motion of the stone can be described using the kinematic equation:
\[
y = v_{0y} t + \frac{1}{2} a t^2
\]
where:
- \(y\) is the vertical displacement (which will be negative, as the stone is falling down).
- \(v_{0y}\) is the initial vertical velocity component.
- \(t = 3.75\) s is the time of flight.
- \(a = -9.8\) m/s² is the acceleration due to gravity (negative because it acts downward).

We are interested in finding the height \(h\) of the cliff. The vertical displacement \(y\) from the point of projection to the ground is:
\[
y = v_{0y} t + \frac{1}{2} a t^2
\]

Since the stone falls down from the initial height \(h\), we have:
\[
-h = v_{0y} t + \frac{1}{2} a t^2
\]
So, the height of the cliff \(h\) can be calculated as:
\[
h = -\left( v_{0y} t + \frac{1}{2} a t^2 \right)
\]

### Step 3: Calculation
Let's calculate the height step by step.

1. Calculate the vertical component of the initial velocity:
   \[
   v_{0y} = 25.0 \, \text{m/s} \cdot \sin(40.0^\circ)
   \]

2. Use the kinematic equation to find \(h\):
   \[
   h = -\left( v_{0y} \cdot 3.75 \, \text{s} + \frac{1}{2} \cdot (-9.8 \, \text{m/s}^2) \cdot (3.75 \, \text{s})^2 \right)
   \]

Let's compute this.The height of the cliff is approximately **8.64 meters**.

Would you like more details or have any further questions?

Here are 5 related questions you might find interesting:
1. How far horizontally did the stone travel before hitting the ground?
2. What was the final velocity of the stone just before impact?
3. How would the height change if the angle was increased to 50°?
4. What would be the time of flight if the stone was thrown vertically downward?
5. How would air resistance affect the height calculation?

**Tip:** Breaking down vectors into their components simplifies solving problems involving angles.

Learn how to calculate the height of a cliff using projectile motion and kinematic equations. Given the initial velocity and angle of projection, find the height of the cliff where the stone hits the ground after 3.75 seconds. This problem involves breaking down velocity vectors, applying kinematic equations, and understanding the effects of gravity on vertical motion. Suitable for advanced high school students and physics enthusiasts.

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