When a ball is thrown up vertically with velocity `v_0`, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity

Options

A. `sqrt(3)v_(0)`
B. `3 v_(0)`
C. `9 v_(0)`
D. `3//2 v_(0)`

Explanation

To solve the problem, we need to determine the initial velocity required to triple the maximum height reached by a ball thrown vertically upward.

### Step-by-Step Solution:

1. **Understand the relationship between initial velocity and maximum height**:
The maximum height \( h \) reached by a ball thrown vertically upwards with an initial velocity \( v_0 \) can be derived from the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity (0 m/s at maximum height),
- \( u \) = initial velocity (\( v_0 \)),
- \( a \) = acceleration (which is \(-g\) due to gravity, but we will consider it as \( g \) for height calculations),
- \( s \) = maximum height (\( h \)).

2. **Set up the equation for maximum height**:
At maximum height, the final velocity \( v = 0 \). Thus, the equation becomes:
\[ 0 = v_0^2 - 2gh \]
Rearranging gives:
\[ v_0^2 = 2gh \]

3. **Determine the new height when tripled**:
If we want to triple the maximum height, the new height \( h' \) will be:
\[ h' = 3h \]

4. **Set up the equation for the new height**:
Using the same kinematic equation for the new height, we have:
\[ v'^2 = 2g(3h) \]
where \( v' \) is the new initial velocity required to reach the new height.

5. **Relate the new initial velocity to the original**:
Substituting \( h' \) into the equation gives:
\[ v'^2 = 6gh \]
Now, we can substitute \( h \) from the earlier equation \( h = \frac{v_0^2}{2g} \):
\[ v'^2 = 6g\left(\frac{v_0^2}{2g}\right) \]
Simplifying this:
\[ v'^2 = 3v_0^2 \]

6. **Solve for the new initial velocity**:
Taking the square root of both sides gives:
\[ v' = \sqrt{3}v_0 \]

### Final Answer:
To triple the maximum height, the ball should be thrown with an initial velocity of \( \sqrt{3}v_0 \).