A frog sits on the end pf a long boord of length L. the boord rests on a fricationless horizontal table. The frog wants os the minimum takes - off speed i.e relative to ground v that allows the frog yo do the trick? The board and the frog have equal masses.

Explanation

To solve the problem of determining the minimum take-off speed of the frog relative to the ground, we can follow these steps:

### Step 1: Understand the System
We have a frog sitting at one end of a long board of length \( L \). The board is resting on a frictionless horizontal table, and both the frog and the board have equal masses.

### Step 2: Apply Conservation of Momentum
Since there are no external horizontal forces acting on the system (frog + board), we can apply the principle of conservation of momentum. Let \( m \) be the mass of the frog (which is equal to the mass of the board).

When the frog jumps off the board with a velocity \( v \) relative to the ground, the board will move in the opposite direction. Let the velocity of the board after the frog jumps be \( u \).

According to the conservation of momentum:
\[ mv + mu = 0 \]
This simplifies to:
\[ mv = -mu \]
From this, we can express \( u \) in terms of \( v \):
\[ u = -\frac{v}{1} = -v \]

### Step 3: Relate Distances Covered
The frog jumps a horizontal distance \( x \) while the board moves a distance \( L - x \) in the opposite direction. The time \( t \) taken for both the frog and the board to cover their respective distances can be expressed as:
\[ t = \frac{x}{v} \quad \text{(for the frog)} \]
\[ t = \frac{L - x}{u} \quad \text{(for the board)} \]
Since both events happen simultaneously, we can set these two equations equal:
\[ \frac{x}{v} = \frac{L - x}{u} \]

### Step 4: Substitute for \( u \)
Substituting \( u = -\frac{v}{1} \) into the equation gives:
\[ \frac{x}{v} = \frac{L - x}{-v} \]
Cross-multiplying leads to:
\[ x(-v) = v(L - x) \]
This simplifies to:
\[ -xv = vL - vx \]
Rearranging gives:
\[ 2xv = vL \]
Thus, we find:
\[ x = \frac{L}{2} \]

### Step 5: Use Kinematic Equations
Now, we can use the kinematic equations to relate the distance \( x \) to the velocity \( v \). The horizontal distance covered by the frog can also be expressed as:
\[ x = v \cos(\theta) t \]
And for the vertical motion:
\[ x = \frac{v^2 \sin(2\theta)}{g} \]
Setting these equal gives:
\[ \frac{L}{2} = \frac{v^2 \sin(2\theta)}{g} \]

### Step 6: Solve for \( v \)
Rearranging the equation yields:
\[ v^2 = \frac{gL}{2 \sin(2\theta)} \]
To find the minimum value of \( v \), we need to maximize \( \sin(2\theta) \). The maximum value of \( \sin(2\theta) \) is 1, which occurs when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \).

Thus, we have:
\[ v^2 = \frac{gL}{2} \]
Taking the square root gives:
\[ v = \sqrt{\frac{gL}{2}} \]

### Step 7: Substitute \( g \)
Assuming \( g = 10 \, \text{m/s}^2 \):
\[ v = \sqrt{\frac{10L}{2}} = \sqrt{5L} \]

### Final Answer
The minimum take-off speed \( v \) that allows the frog to jump off the board is:
\[ v = \sqrt{5L} \, \text{m/s} \]