A stone is thrown outward from the top of a 59.4 m high cliff with an upward velocity component of 19.5 m/s. How long is the stone in the air? Assume that it lands on the ground below the cliff, and that the ground below the cliff is flat.

To determine how long the stone is in the air, we can break down the motion into two components: vertical and horizontal. The time the stone spends in the air is determined by its vertical motion, as the horizontal motion does not affect the time of flight.

### Step 1: Vertical Motion Analysis

The vertical motion can be described by the following kinematic equation:

\[
y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2
\]

where:
- \( y(t) \) is the vertical position of the stone at time \( t \),
- \( y_0 = 59.4 \) m is the initial height of the stone,
- \( v_{0y} = 19.5 \) m/s is the initial vertical velocity component,
- \( g = 9.8 \) m/s² is the acceleration due to gravity (acting downward),
- \( t \) is the time in seconds.

When the stone hits the ground, \( y(t) = 0 \). Thus, the equation becomes:

\[
0 = 59.4 + 19.5t - \frac{1}{2} \times 9.8 \times t^2
\]

Simplifying this equation:

\[
0 = 59.4 + 19.5t - 4.9t^2
\]

### Step 2: Solve the Quadratic Equation

This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where:
- \( a = -4.9 \),
- \( b = 19.5 \),
- \( c = 59.4 \).

We can use the quadratic formula to solve for \( t \):

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substituting the values:

\[
t = \frac{-19.5 \pm \sqrt{(19.5)^2 - 4(-4.9)(59.4)}}{2(-4.9)}
\]

Let's calculate this.

The solutions to the quadratic equation are approximately \( t_1 = -2.02 \) seconds and \( t_2 = 6.00 \) seconds. 

Since time cannot be negative, we discard the negative value. Thus, the stone is in the air for approximately **6.00 seconds**.

Would you like more details or have any questions? Here are some related questions to consider:

1. What would happen if the stone were thrown with a different initial vertical velocity?
2. How would the time in the air change if the cliff were taller or shorter?
3. What is the maximum height the stone reaches during its flight?
4. How far horizontally does the stone travel while in the air?
5. How would the problem change if there was air resistance?

**Tip:** When solving problems involving projectile motion, breaking down the motion into vertical and horizontal components simplifies the analysis.

Learn how to calculate the time a stone spends in the air after being thrown outward from a 59.4 m high cliff with an initial vertical velocity of 19.5 m/s. This problem involves projectile motion and uses kinematic equations along with solving a quadratic equation. Suitable for students in Grades 10-12.

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