A stone is launched straight up by a slingshot with an initial speed of 19.6 m/s. The stone is located 1.50 m above the ground when launched. Find the total time the stone is in the air before hitting the ground.

To find the total time the stone is in the air before hitting the ground, we can use the kinematic equations for projectile motion. 

### Step-by-Step Solution

1. **Identify the given values:**
   - Initial velocity (\(v_0\)) = 19.6 m/s (upward)
   - Initial height (\(h_0\)) = 1.50 m
   - Acceleration due to gravity (\(g\)) = 9.8 m/s² (downward)

2. **Use the kinematic equation for vertical motion:**

   The position of the stone as a function of time (\(t\)) is given by:

   \[
   h(t) = h_0 + v_0 t - \frac{1}{2} g t^2
   \]

   To find the total time the stone is in the air, we need to solve for \(t\) when the stone hits the ground (\(h(t) = 0\)):

   \[
   0 = 1.50 + 19.6 t - \frac{1}{2} \times 9.8 \times t^2
   \]

   Simplifying the equation:

   \[
   0 = 1.50 + 19.6 t - 4.9 t^2
   \]

3. **Solve the quadratic equation:**

   Rearrange into standard form:

   \[
   4.9 t^2 - 19.6 t - 1.5 = 0
   \]

   Use the quadratic formula:

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

   where \(a = 4.9\), \(b = -19.6\), and \(c = -1.5\).

   Plugging in the values:

   \[
   t = \frac{-(-19.6) \pm \sqrt{(-19.6)^2 - 4 \times 4.9 \times (-1.5)}}{2 \times 4.9}
   \]

   Simplify inside the square root:

   \[
   t = \frac{19.6 \pm \sqrt{384.16 + 29.4}}{9.8}
   \]

   \[
   t = \frac{19.6 \pm \sqrt{413.56}}{9.8}
   \]

   \[
   t = \frac{19.6 \pm 20.33}{9.8}
   \]

   This gives two solutions:

   \[
   t_1 = \frac{19.6 + 20.33}{9.8} \approx 4.08 \text{ s}
   \]

   \[
   t_2 = \frac{19.6 - 20.33}{9.8} \approx -0.07 \text{ s}
   \]

   Since time cannot be negative, we take the positive value:

   \[
   t = 4.08 \text{ s}
   \]

**Therefore, the total time the stone is in the air before hitting the ground is approximately 4.08 seconds.**

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. What is the maximum height reached by the stone?
2. How long does it take for the stone to reach its maximum height?
3. What would the total time in the air be if the initial speed were doubled?
4. How would the time in the air change if the stone were launched from a height of 5 meters instead of 1.5 meters?
5. What is the impact speed of the stone when it hits the ground?

**Tip:** When solving projectile motion problems, always clearly separate initial conditions (like initial velocity and height) and ensure consistent use of positive and negative signs to indicate direction (upward vs. downward).

Find out how to calculate the total time a stone is in the air when launched from 1.5 m above the ground with an initial speed of 19.6 m/s. This solution involves kinematic equations and solving a quadratic equation, making it suitable for high school physics students in Grades 9-12.

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