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 upwards.

### Step-by-Step Solution:

1. **Understanding the maximum height formula**:
The maximum height \( h \) reached by an object thrown upwards with an initial velocity \( v_0 \) can be calculated using the formula:
\[ h = \frac{v_0^2}{2g} \]
where \( g \) is the acceleration due to gravity.

2. **Setting up the equation for tripling the height**:
If we want to triple the maximum height, the new height \( h' \) will be:
\[ h' = 3h \]

3. **Substituting the expression for height**:
Using the formula for maximum height, we can express \( h' \) in terms of the new initial velocity \( v' \):
\[ h' = \frac{(v')^2}{2g} \]
Therefore, we can set up the equation:
\[ 3h = \frac{(v')^2}{2g} \]

4. **Substituting the expression for \( h \)**:
From the first equation, we know:
\[ h = \frac{v_0^2}{2g} \]
Substituting this into the equation for \( h' \):
\[ 3 \left(\frac{v_0^2}{2g}\right) = \frac{(v')^2}{2g} \]

5. **Simplifying the equation**:
We can cancel \( \frac{1}{2g} \) from both sides:
\[ 3v_0^2 = (v')^2 \]

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

### Final Answer:
Thus, the initial velocity required to triple the maximum height is:
\[ v' = \sqrt{3} v_0 \]