A person can throw a stone to a maximum height of h meter. The maximum distance to which he can throw the stone is:

Options

A. `(h)/(2)`
B. h
C. 2h
D. 3h

Answer

To find the maximum distance to which a person can throw a stone, we can use the principles of projectile motion. Here’s a step-by-step solution:

### Step 1: Understand the relationship between height and initial velocity
The maximum height \( h \) that the stone can reach is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 2: Determine the angle for maximum height
The maximum height is achieved when the angle \( \theta \) is 90 degrees (i.e., when the stone is thrown straight up). In this case, the formula simplifies to:
\[ h = \frac{u^2}{2g} \]

### Step 3: Rearranging the height equation to find initial velocity
From the equation for height, we can express the initial velocity \( u \) in terms of \( h \):
\[ u^2 = 2gh \]

### Step 4: Use the range formula for projectile motion
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
The range is maximized when \( \theta = 45^\circ \) (i.e., when the stone is thrown at an angle of 45 degrees). In this case, \( \sin 2\theta = \sin 90^\circ = 1 \), so the range simplifies to:
\[ R = \frac{u^2}{g} \]

### Step 5: Substitute the expression for \( u^2 \) into the range formula
Now, substituting \( u^2 = 2gh \) into the range formula:
\[ R = \frac{2gh}{g} \]
This simplifies to:
\[ R = 2h \]

### Conclusion
Thus, the maximum distance to which the person can throw the stone is:
\[ \boxed{2h} \]