You throw the ball from a wndow that is 0.9m above the ground at an angle of 12degree downwards from the horizontal plane. How far from the wall of the house does the ball hit the ground if its initital speed is 12.7 m/s Air resistance is not taken into account 

In this problem, we're given the initial conditions of a projectile motion scenario:

- The ball is thrown from a height of \( 0.9 \, \text{m} \).
- The initial speed of the ball is \( 12.7 \, \text{m/s} \).
- The angle of projection is \( 12^\circ \) downwards from the horizontal.
- We're ignoring air resistance.

We aim to determine how far from the wall (i.e., horizontal distance) the ball hits the ground.

### Step 1: Break the Initial Velocity into Components
Since the ball is thrown at an angle of \( 12^\circ \) below the horizontal, we can decompose the initial velocity \( v_0 = 12.7 \, \text{m/s} \) into horizontal and vertical components:
- Horizontal component: \( v_{0x} = v_0 \cos(12^\circ) \)
- Vertical component: \( v_{0y} = v_0 \sin(12^\circ) \)

### Step 2: Use the Kinematic Equation for Vertical Motion
We know the initial height \( h = 0.9 \, \text{m} \), and the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \). The vertical displacement when the ball hits the ground is \( -0.9 \, \text{m} \), since it falls below the release point.

We can use the following kinematic equation for vertical motion:

\[
y = v_{0y} t + \frac{1}{2} g t^2
\]

where:
- \( y = -0.9 \, \text{m} \) (the change in vertical height),
- \( v_{0y} = -v_0 \sin(12^\circ) \) (since the ball is projected downwards),
- \( g = 9.8 \, \text{m/s}^2 \).

### Step 3: Solve for Time of Flight
Solving this quadratic equation for \( t \), the time it takes for the ball to hit the ground, will allow us to find the horizontal distance.

### Step 4: Calculate the Horizontal Distance
Once we know the time \( t \), we can compute the horizontal distance \( d \) using the horizontal velocity component:

\[
d = v_{0x} t = v_0 \cos(12^\circ) t
\]

Now let's calculate.The ball will hit the ground approximately **9.64 meters** away from the wall of the house.

Would you like more details or have any other questions?

Here are five related questions for further practice:

1. How would the horizontal distance change if the initial speed were increased to 15 m/s?
2. What happens to the time of flight if the ball is thrown from a higher window, say 1.5 meters above the ground?
3. How does air resistance affect the trajectory and distance traveled?
4. If the ball were thrown upwards at 12 degrees instead of downwards, how far would it travel horizontally before hitting the ground?
5. How does the angle of projection affect both the horizontal distance and time of flight?

**Tip:** Always decompose projectile motion problems into horizontal and vertical components to simplify your calculations.

You throw the ball from a window that is 0.9m above the ground at an angle of 12 degrees downwards from the horizontal plane. How far from the wall of the house does the ball hit the ground if its initial speed is 12.7 m/s? Air resistance is not taken into account.

Learn how to calculate the horizontal distance traveled by a ball thrown from a 0.9-meter height at an angle of 12 degrees downwards with an initial speed of 12.7 m/s. This problem involves key concepts in projectile motion, trigonometry, and kinematics, suitable for high school students.

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