A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower’s hand with a speed of 6.00m/s. Air resistance may be ignored, so the water balloon is freefall after it leaves the thrower’s hand. a. What is its speed after falling for 2.00 s? b. How far does it fall in 2.00 s? c. What is the magnitude of its velocity after falling 10.0 m?

To solve this problem, we'll use the **kinematic equations** for uniformly accelerated motion, assuming no air resistance. The acceleration is due to gravity, \( g = 9.81 \, \text{m/s}^2 \), and the motion is in the downward direction.

### Given dataTo solve this problem, we'll use the **kinematic equations** for uniformly accelerated motion, assuming no air resistance. The acceleration is due to gravity, \( g = 9.81 \, \text{m/s}^2 \), and the motion is in the downward direction.

### Given 
- Initial velocity, \( v_0 = 6.00 \, \text{m/s} \) (downward)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)
- Time, \( t = 2.00 \, \text{s} \) for parts (a) and (b)
- Distance, \( d = 10.0 \, \text{m} \) for part (c)

### (a) Speed after falling for 2.00 s:

We use the **velocity equation**:
\[
v = v_0 + g t
\]

Substituting the known values:
\[
v = 6.00 \, \text{m/s} + (9.81 \, \text{m/s}^2)(2.00 \, \text{s})
\]

### (b) Distance fallen in 2.00 s:

We use the **position equation**:
\[
d = v_0 t + \frac{1}{2} g t^2
\]

Substituting the known values:
\[
d = (6.00 \, \text{m/s})(2.00 \, \text{s}) + \frac{1}{2} (9.81 \, \text{m/s}^2)(2.00 \, \text{s})^2
\]

### (c) Velocity after falling 10.0 m:

We use the **third kinematic equation**:
\[
v^2 = v_0^2 + 2 g d
\]

Substituting the known values:
\[
v^2 = (6.00 \, \text{m/s})^2 + 2(9.81 \, \text{m/s}^2)(10.0 \, \text{m})
\]
\[
v = \sqrt{(6.00)^2 + 2(9.81)(10.0)}
\]

Now let's calculate each part.

### (a) Velocity after 2.00 s:

\[
v = 6.00 + (9.81)(2.00) = 6.00 + 19.62 = 25.62 \, \text{m/s}
\]

### (b) Distance fallen in 2.00 s:

\[
d = (6.00)(2.00) + \frac{1}{2} (9.81)(2.00)^2 = 12.00 + \frac{1}{2} (9.81)(4.00) = 12.00 + 19.62 = 31.62 \, \text{m}
\]

### (c) Velocity after falling 10.0 m:

\[
v = \sqrt{(6.00)^2 + 2(9.81)(10.0)} = \sqrt{36.00 + 196.2} = \sqrt{232.2} \approx 15.23 \, \text{m/s}
\]

### Final Answers:
1. **(a)** The speed after 2.00 s is **25.62 m/s**.
2. **(b)** The balloon falls **31.62 m** in 2.00 s.
3. **(c)** The speed after falling 10.0 m is **15.23 m/s**.

Would you like more details or further explanation?

Here are 5 related questions to explore:
1. What would happen to the distance fallen if the initial speed was doubled?
2. How does air resistance affect the final velocity in real-world scenarios?
3. How would the equations change if the balloon was thrown upward instead of downward?
4. What is the impact of changing the acceleration due to gravity on another planet?
5. How can we calculate the maximum height reached if the balloon was thrown upwards?

**Tip:** Always check the direction of motion and the sign of acceleration when solving free-fall problems.

A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower’s hand with a speed of 6.00 m/s. Air resistance may be ignored, so the water balloon is in freefall after it leaves the thrower’s hand. a. What is its speed after falling for 2.00 s? b. How far does it fall in 2.00 s? c. What is the magnitude of its velocity after falling 10.0 m?

Explore how to calculate the speed and distance of a water balloon in freefall after being thrown downward with an initial speed of 6.00 m/s. Using key kinematic equations, this detailed step-by-step solution shows the velocity after 2.00 seconds and after falling 10.0 meters. Suitable for high school students in Grades 9-12.

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