Question

A rock is thrown horizontally from a cliff at 25m/s. What will its horizontal and vertical velocities be after 2.0s?

Answer 1

To find the horizontal and vertical velocities of the rock after 2.0 seconds, we can analyze the motion in both directions separately.

### Horizontal Motion:

1. **Initial Horizontal Velocity (Vx)**: The rock is thrown horizontally at a speed of 25 m/s, and there’s no horizontal acceleration (assuming air resistance is negligible). Therefore, the horizontal velocity remains constant:
\[
V_{x} = 25 \, \text{m/s}
\]

### Vertical Motion:

1. **Initial Vertical Velocity (Vy)**: The initial vertical velocity is 0 m/s because the rock is thrown horizontally.

2. **Vertical Acceleration (ay)**: The only vertical force acting on the rock is gravity, which accelerates the rock downward at approximately \( g = 9.81 \, \text{m/s}^2 \).

3. **Vertical Velocity after 2 seconds (Vy)**: We can calculate the vertical velocity after 2.0 seconds using the following kinematic equation:
\[
V_{y} = V_{iy} + a_{y} t
\]
where:
- \( V_{iy} = 0 \, \text{m/s} \)
- \( a_{y} = 9.81 \, \text{m/s}^2 \)
- \( t = 2.0 \, \text{s} \)

Substituting the values:
\[
V_{y} = 0 + (9.81 \, \text{m/s}^2)(2.0 \, \text{s}) = 19.62 \, \text{m/s}
\]

### Summary of Velocities:

- **Horizontal Velocity (Vx)**: 25 m/s (remains constant)
- **Vertical Velocity (Vy)**: 19.62 m/s (downward)

Thus, after 2.0 seconds, the rock's horizontal velocity is 25 m/s, and its vertical velocity is 19.62 m/s downward.