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## Homework Statement

A frog sits on the end of a long board of length L=5m . The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board . What is the minimum take-off speed (in s.i. Units) i.e.,relative to ground 'v' that allows the frog to do the trick ? The board and the frog have equal masses.

## Homework Equations

## The Attempt at a Solution

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As no external force act in the horizontal direction on the system so horizontal velocity of the frog is equal to the velocity of the plank (their directions are opposite). And also the position of the CoM of the system at final position and initial position should be same.

Now , the CoM of the plank will be displaced by length ##5/2##.

So ##t=\frac{5}{2V_{x}}##..................(1)

In this time period the from will jump and land on the plank.

So ##t=\frac{V_{y}}{5}##..............(2)

Using eq(1) and (2) we get

##V_{y}V_{x}=\frac{25}{2}##........(3)

Let the net velocity of the frog be ##v_{0}##

So ##{ V }_{ 0 }^{ 2 }={ V }_{ x }^{ 2 }+{ V }_{ y }^{ 2 }##..................(4)

So ##{ V }_{ 0 }^{ 2 }={ V }_{ x }^{ 2 }+{ \left( \frac { 25 }{ 2{ V }_{ x } } \right) }^{ 2 }##

##V_{0}## will be minimum when ##V_{0} ^{2}## will be minimum.

So differentiating the equation wrt ##V_{x}## and equating to 0 I got that this function would have minimum value when ##V_{x}=\frac { 5 }{ \sqrt { 2 } } ##. So ##V_{y}=\frac { 5 }{ \sqrt { 2 } } ##

Putting this in equation(4) I got minimum value of ##v_{0}=5m/s##.

But ##5m/s## is not in the options provided. I am doubting my solution. Am I right?