If Emily throws a ball at an angle of  30° below the horizonal with a speed of 14 m/s how far from the base of the dorm should Allison stand to catch the ball? Assume the vertical distance between where Emily releases the ball and Allison catches it is 4.0 m

To solve this problem, we will break it into two components: horizontal and vertical motion.

### Known values:
- Initial speed, \( v_0 = 14 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \) (below the horizontal)
- Vertical displacement, \( \Delta y = -4.0 \, \text{m} \) (since the ball is moving downward)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

### Step 1: Horizontal and Vertical Components of Initial Velocity
The initial velocity can be split into two components:
- **Horizontal velocity**, \( v_{0x} = v_0 \cos(\theta) \)
- **Vertical velocity**, \( v_{0y} = v_0 \sin(\theta) \)

\[
v_{0x} = 14 \cos(30^\circ)
\]
\[
v_{0y} = 14 \sin(30^\circ)
\]

### Step 2: Time of Flight
Next, we use the kinematic equation for vertical motion to determine the time it takes for the ball to reach Allison:
\[
\Delta y = v_{0y} t + \frac{1}{2} g t^2
\]
Substitute the known values and solve for \( t \). Since the angle is below the horizontal, \( v_{0y} \) will be negative:
\[
-4.0 = -14 \sin(30^\circ) t + \frac{1}{2} (9.8) t^2
\]

### Step 3: Horizontal Distance
Once we have the time of flight \( t \), the horizontal distance can be calculated using:
\[
x = v_{0x} t
\]

Let me calculate this now.It seems that the time equation yielded a complex solution, which typically happens when there’s a discrepancy in interpreting the physical scenario. I'll review the kinematic equation and ensure it properly handles the direction and magnitude of the velocities. Let me reassess and resolve this step carefully.It seems there was an issue identifying the correct time of flight from the solutions. I'll try solving the quadratic equation for time more carefully and ensure that I isolate the physically meaningful solution. Let's correct that now.The time of flight is approximately \( 1.87 \, \text{seconds} \), and the horizontal distance is about \( 22.62 \, \text{meters} \). 

This means Allison should stand about **22.6 meters** away from the base of the dorm to catch the ball.

Would you like a more detailed breakdown or have any other questions?

Here are 5 related questions to explore further:
1. How would the distance change if the initial speed was increased to 16 m/s?
2. What if the angle of projection was 45° instead of 30° below the horizontal?
3. How would the result differ if the ball was thrown on a different planet with a different gravity value?
4. What happens if there is air resistance?
5. How long would the ball stay in the air if it was thrown horizontally?

**Tip**: Splitting velocity into horizontal and vertical components is key for projectile motion problems!

If Emily throws a ball at an angle of 30° below the horizontal with a speed of 14 m/s, how far from the base of the dorm should Allison stand to catch the ball? Assume the vertical distance between where Emily releases the ball and Allison catches it is 4.0 m.

Solve a physics problem involving projectile motion where Emily throws a ball at 30° below the horizontal with a speed of 14 m/s. Learn how to calculate the horizontal distance Allison should stand from the base of the dorm, given a vertical displacement of 4.0 m. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation