You throw a ball straight up with an initial speed of 10 m/s. Take upward direction as + Y axis and `g = 10 m//s^(2)` :-

Options

A. speed of paritcle increases
B. speed of particle decreases
C. Acceleration of particle increase
D. Acceleration of particle decreases

Answer

To solve the problem of throwing a ball straight up with an initial speed of 10 m/s, we will analyze the motion of the ball under the influence of gravity.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Initial speed (u) = 10 m/s (upward)
- Acceleration due to gravity (g) = -10 m/s² (downward)

2. **Use the Equation of Motion**:
We can use the equation of motion to find the maximum height (h) reached by the ball:
\[ v^2 = u^2 + 2as \]
where:
- \(v\) = final velocity (0 m/s at the maximum height)
- \(u\) = initial velocity (10 m/s)
- \(a\) = acceleration (-10 m/s²)
- \(s\) = displacement (height, h)

3. **Substituting the Values**:
At the maximum height, the final velocity \(v = 0\). Substituting the values into the equation:
\[ 0 = (10)^2 + 2(-10)(h) \]
Simplifying this gives:
\[ 0 = 100 - 20h \]

4. **Solve for h**:
Rearranging the equation:
\[ 20h = 100 \]
\[ h = \frac{100}{20} = 5 \text{ m} \]

5. **Conclusion**:
The maximum height reached by the ball is 5 meters.