A projectile is thrown from the top of a 1.43 m high table at an angle 28 degrees and lands 5.48 m away horizontally. What is the init speed

To find the initial speed of the projectile, we need to apply the equations of projectile motion. The key idea is to break the motion into horizontal and vertical components.

### Step 1: Break into components
Given dataTo find the initial speed of the projectile, we need to apply the equations of projectile motion. The key idea is to break the motion into horizontal and vertical components.

### Step 1: Break into components
Given 
- Height of the table: \( h = 1.43 \, \text{m} \)
- Angle of projection: \( \theta = 28^\circ \)
- Horizontal distance (range): \( R = 5.48 \, \text{m} \)
- Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \)

We need to determine the initial speed \( v_0 \).

### Step 2: Horizontal motion equation
The horizontal distance covered is given by:
\[
R = v_0 \cos(\theta) \cdot t
\]
where \( t \) is the time of flight.

### Step 3: Vertical motion equation
For vertical motion, the vertical displacement \( \Delta y = -1.43 \, \text{m} \) (since the projectile falls below its starting height) is given by:
\[
\Delta y = v_0 \sin(\theta) \cdot t - \frac{1}{2} g t^2
\]

Substitute \( \Delta y = -1.43 \, \text{m} \) and solve the system of equations to find the initial speed \( v_0 \).

### Step 4: Solving for time \( t \)
From the horizontal equation:
\[
t = \frac{R}{v_0 \cos(\theta)}
\]

Substitute this into the vertical equation:
\[
-1.43 = v_0 \sin(\theta) \cdot \frac{R}{v_0 \cos(\theta)} - \frac{1}{2} g \left( \frac{R}{v_0 \cos(\theta)} \right)^2
\]

Now, solve this equation for \( v_0 \).

Let's compute this step-by-step.The positive solution for the initial speed \( v_0 \) is approximately \( 6.59 \, \text{m/s} \). 

Thus, the initial speed of the projectile is about **6.59 m/s**.

Do you want more details or have any questions?

Here are 5 related questions for further practice:
1. What would be the time of flight for the projectile?
2. How would the initial speed change if the angle of projection were increased?
3. What is the maximum height the projectile reaches during its flight?
4. How does the range of the projectile change if the initial speed is doubled?
5. What would happen to the range if the table height were halved?

**Tip:** In projectile motion, breaking the motion into horizontal and vertical components simplifies complex problems.

A projectile is thrown from the top of a 1.43 m high table at an angle 28 degrees and lands 5.48 m away horizontally. What is the initial speed?

Learn how to calculate the initial speed of a projectile thrown at an angle of 28 degrees from a height of 1.43 m, covering a horizontal distance of 5.48 m. This problem demonstrates the use of projectile motion equations and trigonometric principles. Suitable for Grades 10-12.

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