You flick a ball from a tabletop, so that the ball lands on the floor 1.20 m below. The ball's initial speed is 2.10 m/s. Use g = 9.80 m/s2.
(a) If the ball is launched off the table horizontally, what is the horizontal distance between the launch point and the point where the ball hits the floor?
1.04
 
Correct: Your answer is correct.
 m

(b) If the ball is launched off the table at a 30° angle above the horizontal, what is the horizontal distance between the launch point and the point where the ball hits the floor?

To solve part (b), where the ball is launched off the table at a 30° angle above the horizontal, we'll analyze the motion in two dimensions: horizontal and vertical. The given parameters are:

- Initial speed (\(v_0\)) = 2.10 m/s
- Launch angle (\(\theta\)) = 30°
- Height (\(h\)) = 1.20 m
- Acceleration due to gravity (\(g\)) = 9.80 m/s\(^2\)

### Step 1: Decompose the initial velocity

The initial velocity can be split into horizontal and vertical components:

\[
v_{0x} = v_0 \cos \theta = 2.10 \cdot \cos 30^\circ
\]

\[
v_{0y} = v_0 \sin \theta = 2.10 \cdot \sin 30^\circ
\]

### Step 2: Calculate the time of flight

The time of flight can be found using the vertical motion equations. Since the ball starts at a height \(h\) and falls to the ground, we can use:

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

Substituting \( y = -1.20 \, \text{m} \) (since it falls 1.20 m below its initial position):

\[
-1.20 = (2.10 \cdot \sin 30^\circ) t - \frac{1}{2} (9.80) t^2
\]

Simplifying the equation:

\[
-1.20 = (2.10 \cdot 0.5) t - 4.90 t^2
\]

\[
-1.20 = 1.05 t - 4.90 t^2
\]

Rearrange into a standard quadratic form:

\[
4.90 t^2 - 1.05 t - 1.20 = 0
\]

Now solve this quadratic equation using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 4.90 \), \( b = -1.05 \), and \( c = -1.20 \).

### Step 3: Calculate the time

Let's compute the time values:The possible times calculated are:

- \( t_1 = 0.613 \, \text{s} \) (positive, valid time)
- \( t_2 = -0.399 \, \text{s} \) (negative, not physically meaningful)

So, the time of flight is \( t = 0.613 \, \text{s} \).

### Step 4: Calculate the horizontal distance

Now, using the horizontal component of the velocity:

\[
d = v_{0x} \cdot t = (2.10 \cdot \cos 30^\circ) \times 0.613
\]

Let’s compute the horizontal distance:The horizontal distance between the launch point and the point where the ball hits the floor is approximately:

\[
\boxed{1.12 \, \text{m}}
\]

Would you like further details, or do you have any questions?

### Related Questions:
1. How would the distance change if the initial speed were increased?
2. What would be the time of flight if the launch angle were 45°?
3. How does the launch angle affect the horizontal distance?
4. What would happen to the trajectory if air resistance were considered?
5. Can you calculate the maximum height reached if launched at 30°?

**Tip:** Breaking down initial velocity into horizontal and vertical components helps in analyzing projectile motion accurately.

You flick a ball from a tabletop, so that the ball lands on the floor 1.20 m below. The ball's initial speed is 2.10 m/s. Use g = 9.80 m/s2. (a) If the ball is launched off the table horizontally, what is the horizontal distance between the launch point and the point where the ball hits the floor? (b) If the ball is launched off the table at a 30° angle above the horizontal, what is the horizontal distance between the launch point and the point where the ball hits the floor?

Solve a projectile motion problem where a ball is flicked from a tabletop with an initial speed of 2.10 m/s, falling 1.20 meters below. Learn how to calculate the horizontal distance when the ball is launched at a 30° angle above the horizontal. Key concepts include trigonometry, quadratic equations, and projectile motion. Suitable for Grades 10-12 physics students.

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