While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.19 m/s. The stone subsequently falls to the ground, which is 16.9 m below the point where the stone leaves your hand. At which speed does the stone impact the ground? How much time is the stone in the air (ignore air resistance)?

To solve this problem, we need to find two things:

1. The speed of the stone when it impacts the ground.
2. The time the stone is in the air.

We'll use the following kinematic equations to solve for the final speed and time of flight, assuming downward is the positive direction:

1. **Final velocity equation**:
\[
v^2 = v_0^2 + 2g h
\]

2. **Time of flight equation**:
\[
h = v_0 t + \frac{1}{2} g t^2
\]

where:
- \( v \) is the final velocity of the stone when it impacts the ground.
- \( v_0 = -6.19 \, \text{m/s} \) is the initial velocity (negative because it's upward).
- \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity (positive in the downward direction).
- \( h = 16.9 \, \text{m} \) is the total distance the stone falls (positive because it's downward).
- \( t \) is the time in seconds.

### Step 1: Calculate the Final Velocity

Let's plug in the values to compute the final velocity \( v \):

\[
v^2 = (-6.19)^2 + 2 \times 9.81 \times 16.9
\]

### Step 2: Calculate the Time of Flight

Now, we can solve for the time of flight using the quadratic formula:

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

Solving this will give us two solutions for \( t \), one of which will be positive and represent the actual time the stone is in the air.

Let's compute these two values.The stone impacts the ground with a speed of approximately **19.23 m/s**.

The time the stone is in the air is approximately **2.59 seconds**.

Would you like a more detailed explanation, or do you have any other questions?

Here are some related questions you might be interested in:

1. How would air resistance affect the time and speed of the stone?
2. What is the maximum height the stone reaches above the point of release?
3. How would the initial speed change if the stone were thrown downward instead?
4. What if the building was 50% taller; how would that affect the final speed?
5. How do we derive the kinematic equations used in this solution?

**Tip:** When solving kinematic problems, always establish a consistent sign convention for directions to avoid errors in calculations.

Learn how to calculate the speed of a stone upon impact and the time it stays in the air after being thrown upward from a building. This kinematics problem uses key formulas such as v^2 = v_0^2 + 2gh and h = v_0t + (1/2)gt^2 to solve the motion of the stone under gravity. Suitable for high school physics students.

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