The velocity at the midway point of a ball able to reach a height y when thrown with velocity v, at the origin is

Options

A. `v_(i)//2`
B. `sqrt(v_(1)2g y)`
C. `sqrt(v_(i)^(2)//2)`
D. `sqrt(v_(i)^(2)//2 g y)`

Answer

To find the velocity at the midway point of a ball that can reach a height \( h \) when thrown with an initial velocity \( v_0 \) from the origin, we can follow these steps:

### Step 1: Determine the maximum height
The maximum height \( H \) reached by the ball can be calculated using the formula:
\[ H = \frac{v_0^2 \sin^2 \theta}{2g} \]
For the case where the ball is thrown vertically upwards, \( \theta = 90^\circ \) and \( \sin^2 \theta = 1 \). Thus, the formula simplifies to:
\[ H = \frac{v_0^2}{2g} \]

### Step 2: Identify the midway height
The midway height \( h \) is half of the maximum height:
\[ h = \frac{H}{2} = \frac{v_0^2}{4g} \]

### Step 3: Use the kinematic equation to find velocity at height \( h \)
Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity at height \( h \),
- \( u \) is the initial velocity \( v_0 \),
- \( a \) is the acceleration due to gravity (which is negative, so \( a = -g \)),
- \( s \) is the distance traveled (which is \( h \)).

Substituting the values:
\[ v^2 = v_0^2 - 2gh \]

### Step 4: Substitute \( h \) into the equation
Now substituting \( h = \frac{v_0^2}{4g} \) into the equation:
\[ v^2 = v_0^2 - 2g \left(\frac{v_0^2}{4g}\right) \]
This simplifies to:
\[ v^2 = v_0^2 - \frac{v_0^2}{2} = \frac{v_0^2}{2} \]

### Step 5: Solve for \( v \)
Taking the square root of both sides gives:
\[ v = \frac{v_0}{\sqrt{2}} \quad \text{or} \quad v \approx 0.707 v_0 \]

### Final Answer
The velocity at the midway point of the ball is:
\[ v = \frac{v_0}{\sqrt{2}} \]

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