2. A rock is thrown at a window that is located 18.0 m above the
ground. The rock is thrown at an angle of 40.0° above horizontal. The
rock is thrown from a height of 2.00 m above the ground with a speed
of 30.0 m/s and experiences no appreciable air resistance. If the rock
strikes the window on its upward trajectory, from what horizontal
distance from the window was it released?

This is a projectile motion problem. To solve for the horizontal distance from the window where the rock was released, we'll use the following key equations of projectile motion:

### Key Variables:
- Initial speed of the rock: \( v_0 = 30.0 \, \text{m/s} \)
- Launch angle: \( \theta = 40.0^\circ \)
- Height of the window: \( y_f = 18.0 \, \text{m} \)
- Height from which the rock is thrown: \( y_0 = 2.00 \, \text{m} \)
- Gravitational acceleration: \( g = 9.81 \, \text{m/s}^2 \)

We will break the motion into horizontal and vertical components:

### 1. Horizontal Motion:
The horizontal component of the initial velocity is:
\[
v_{0x} = v_0 \cos \theta
\]

The horizontal distance traveled is:
\[
x = v_{0x} t
\]
where \( t \) is the time of flight.

### 2. Vertical Motion:
The vertical component of the initial velocity is:
\[
v_{0y} = v_0 \sin \theta
\]

The equation for the vertical displacement is:
\[
y_f = y_0 + v_{0y} t - \frac{1}{2} g t^2
\]
where \( y_f - y_0 \) is the vertical distance between the release point and the window. This simplifies to:
\[
18.0 \, \text{m} = 2.00 \, \text{m} + (30.0 \sin 40^\circ) t - \frac{1}{2} (9.81) t^2
\]
This is a quadratic equation in \( t \), which we will solve for the time \( t \), and then use that value to find the horizontal distance \( x \).

Let me solve for \( t \).The two possible time values obtained are:

- \( t_1 = 1.19 \, \text{seconds} \)
- \( t_2 = 2.74 \, \text{seconds} \)

Since the problem specifies that the rock strikes the window on its upward trajectory, we select the smaller time value, \( t_1 = 1.19 \, \text{seconds} \).

Now, let's use this time to calculate the horizontal distance from where the rock was released:

\[
x = v_{0x} t_1
\]
where \( v_{0x} = v_0 \cos \theta \).

Let me calculate the horizontal distance.It seems like I can’t do more advanced data analysis right now. Please try again later.

If you'd like, I can guide you on how to calculate the final step manually. Would you like me to walk you through it?

Here are some questions related to the problem for further exploration:
1. How would air resistance affect the result?
2. What would happen if the window were at a lower height?
3. How does changing the launch angle affect the horizontal distance?
4. How could we find the speed of the rock when it hits the window?
5. How would the result change if the rock was thrown from a higher point?

**Tip:** Always check for units' consistency in projectile motion problems, ensuring distances are in meters and speeds in meters per second.

A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal. The rock is thrown from a height of 2.00 m above the ground with a speed of 30.0 m/s and experiences no appreciable air resistance. If the rock strikes the window on its upward trajectory, from what horizontal distance from the window was it released?

Learn how to solve a projectile motion problem where a rock is thrown at a 40.0° angle toward a window 18.0 m high, from a height of 2.00 m. We calculate the horizontal distance from the release point using key kinematic equations and the quadratic formula. Suitable for high school physics students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation