A golf ball is launched off the side of a cliff at 10m/s.  If it takes 5 seconds to fall, what is the range of the golf ball?


To solve for the range (horizontal distance) of the golf ball, we will use the following approach:

1. **Horizontal Motion**: The horizontal velocity (\(v_x\)) remains constant because there are no forces acting in the horizontal direction (assuming air resistance is negligible). The horizontal range can be calculated using the formula:
   
   \[
   \text{Range} = v_x \times t
   \]
   where:
   - \(v_x = 10 \, \text{m/s}\) (the horizontal velocity),
   - \(t = 5 \, \text{seconds}\) (the time of flight, which is given by the time it takes to fall).

2. **Calculating the Range**:
   \[
   \text{Range} = 10 \, \text{m/s} \times 5 \, \text{seconds} = 50 \, \text{meters}
   \]

Thus, the range of the golf ball is **50 meters**.

Would you like further details on the vertical motion or any clarifications? 

Here are some related questions you might want to explore:
1. How high is the cliff if the ball takes 5 seconds to fall?
2. What would happen if the launch velocity was doubled?
3. How would the range be affected if the golf ball was launched upward at an angle?
4. What is the effect of air resistance on the golf ball's range?
5. How does the mass of the golf ball affect its motion in this scenario?

**Tip:** In projectile motion, the horizontal and vertical motions are independent of each other.

A golf ball is launched off the side of a cliff at 10m/s. If it takes 5 seconds to fall, what is the range of the golf ball?

Learn how to determine the range of a golf ball launched at 10 m/s from a cliff, using kinematics principles. This problem covers key concepts in projectile motion and horizontal distance calculation, suitable for students in Grades 9-11.

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