A cannonball is launched horizontally from the top of a 123 m cliff and lands 158 m from the base of the cliff determine the original speed in m/s of the cannonball

To determine the original speed of the cannonball, we need to break the problem into horizontal and vertical components.

### Step 1: Vertical Motion (Free Fall)
Since the cannonball is launched horizontally, its initial vertical velocity is \(0 \, \text{m/s}\). The only force acting vertically is gravity, and we can use the following kinematic equation to find the time \(t\) it takes for the cannonball to hit the ground:

\[
y = \frac{1}{2} g t^2
\]

Where:
- \(y = 123 \, \text{m}\) (height of the cliff),
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity),
- \(t\) is the time the cannonball takes to hit the ground.

Solving for \(t\):

\[
123 = \frac{1}{2} \times 9.8 \times t^2
\]
\[
t^2 = \frac{123 \times 2}{9.8}
\]
\[
t^2 = 25.1
\]
\[
t = \sqrt{25.1} \approx 5.01 \, \text{seconds}
\]

### Step 2: Horizontal Motion
Now that we know the time \(t\), we can find the horizontal velocity of the cannonball. The horizontal distance the cannonball travels is given as \(x = 158 \, \text{m}\). Since the horizontal motion is uniform (no acceleration), we use the equation:

\[
x = v_x \cdot t
\]

Where:
- \(x = 158 \, \text{m}\),
- \(v_x\) is the horizontal velocity (which is the original speed since it was launched horizontally),
- \(t = 5.01 \, \text{seconds}\).

Solving for \(v_x\):

\[
v_x = \frac{158}{5.01} \approx 31.54 \, \text{m/s}
\]

### Final Answer:
The original speed of the cannonball is approximately \(31.54 \, \text{m/s}\).

Would you like further clarification or details on any of these steps?

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Here are 5 questions for further practice:
1. If the height of the cliff was doubled, how would that affect the time of flight?
2. What would happen to the horizontal distance if the speed was doubled?
3. How does air resistance affect the horizontal range in real-life scenarios?
4. If the cannonball was launched at an angle, how would the calculations change?
5. What is the significance of neglecting air resistance in this type of problem?

**Tip:** In projectile motion, always treat horizontal and vertical components separately, as they are independent of each other.

A cannonball is launched horizontally from the top of a 123 m cliff and lands 158 m from the base of the cliff. Determine the original speed in m/s of the cannonball.

Learn how to calculate the original speed of a cannonball launched horizontally from a 123-meter cliff that lands 158 meters away from the base. This problem demonstrates the concepts of projectile motion, kinematic equations, and the independence of horizontal and vertical motion. Suitable for Grades 9-12.

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