Frogs don't know physics, do they?

The first thing to notice about this problem is that the overall momentum in the $\hat{x}$ -direction of the frog and board will be conserved throughout the process. Because they have equal mass (the frog and board), and they start off at rest, we can simply write

$v_\text{frog}\cos\theta + v_\text{board} = 0$

The $\cos\theta$ reflects the fact that the frog is not jumping purely sideways but at an angle $\theta$ relative to the table so that he goes up in the air in a projectile motion. By the end of his jump, we want the frog to be on the opposite end of the board. Thus, we can choose coordinates where the frog begins at the origin, and the end of the board starts at $x=L$ .

The motion of the board (after the frog jumps) is given by $\displaystyle x_\text{board} = L - v_\text{board}t$ . The motion of the frog is described by

$\begin{aligned} x_\text{frog} &= v_\text{frog}\cos\theta t \\ y_\text{frog} &= v_\text{frog}\sin\theta t - \frac12 gt^2 \end{aligned}$

When the frog lands, we want the $x$ position of the frog and the end of the board to be the same (i.e. $x_\text{frog}=x_\text{board}$ ), thus the time of coincidence is given by $\displaystyle t^* = \frac{L}{2v_\text{frog}\cos\theta}$ .

We're also constrained by the vertical motion of the frog, which demands that $\displaystyle t^* = 2v_\text{frog}\sin\theta/g$ . Putting these results together, we find

$\displaystyle v = \frac12 \sqrt{\frac{gL}{\cos\theta\sin\theta}}$

This shows that there is no single choice of $v$ or $\theta$ . There are an infinity of $\left(v,\theta\right)$ pairs, which lie along the curves defined by the equality above, that result in the successful performance of the trick. However, we can choose the pair with the smallest value of $v$ such that the trick successfully obtains.

The only free parameter here is in the choice of $\theta$ , the frog's takeoff angle. Plotting this expression as a function of $v$ , it is clear that the velocity hits a minimium at $\theta = \pi/4$

The requisite velocity for this takeoff angle is simply $v = \sqrt{\frac{gL}{2}}$ .

How can you show the minimization condition mathematically?

Let the horizontal component of jump velocity of the frog be x , and the vertical component be y . Conservation of momentum (in horizontal direction) implies that the log will move with speed x in the opposite direction. [ Note that the momentum is not conserved in vertical direction as there is a normal force acting on the frog-planck system ]

The time taken for the frog to land back on the plank = $\frac{2y}{g}$ . And relative to the plank, the speed of the frog is 2x , so the distance it covers is $\frac{4xy}{g}$ = L (the length of the planck)

So, now we know that: xy = $\frac{Lg}{4}$

We have to minimize $(x^2 + y^2)^\frac{1}{2}$ .

Substitute $y = \frac{Lg}{4x}$ in this. Differentiate wrt x. Equate it to zero to minimize, to get:

$x=\frac{\sqrt{Lg}}{2}$ and $y=\frac{\sqrt{Lg}}{2}$

So the angle is 45. and the jump speed is $\sqrt(\frac{Lg}{2}) = \sqrt(\frac{5g}{2})$ = 4.94

Let $A\equiv F(0)\equiv O$ be the initial position of the frog and the first point of the board, and $B$ the second point of the board. Let $(O,x,y)$ be a system of cartesian coordinates. The consevation of the momentum implies: $x_B = L - x_F;\,(1)$ The law of motion if the frog has a takeoff velocity and angle $v_0,{\theta}_0$ leads to:

$x_F=v_0\cos{\theta}_0 t;\,y_F=v_0\sin{\theta}_0 t - \frac {1} {2} g t^2;\,(2)$

The maximum value of $x_F$ is for ${\theta}_0 = \frac \pi 4,\,y=0;\,(3)$ and implies the minimum velocity $v_0$ ; Must be:

$x_F \geq x_B;$ and for $(1)$ $x_F \geq \frac L 2;\,(4)$

From $(3)$ $t^* =\frac{2 v_0 \sin{\theta}_0}{g}=\frac{\sqrt{2}v_0}{g};\,(5)$ the flying time;

From $(2),(3),(4),(5)\,$ we obtain:

$v^2_0 \geq \frac {gL} 2;\,v_0 \geq \sqrt{\frac{gL}{2}}=\sqrt{\frac{9.8 \times 5}{2}}\approx \boxed {4.95 \rm {\frac m s}}$

As their mass is the same and there must be conservation of momentum, the board will go as the same ground velocity than the frog, but in the opposite direction.

That means the frog just have to make a 5m-long jump (horizontally speaking). (The frog would have traveled 5m and the board too in opposite direction, which gives the 10m requested along the board.) Let's then forget about the board and resolve the problem for a 5m jump.

Let's assume that we know the most efficient jump is at 45°. Then vertical speed is the same as the ground speed, and let's call them both $v$ . Let be $t$ the time used to jump.

We have along the vertical axis: $v·t-\frac{1}{2}g·t^2=0$ as the end of the jump is at the same height at the start. This gives $t=0$ (of course, the start is at the same height as the start…) and $t=\frac{2v}{g}$ .

Along the horizontal axis: $v·t=5 \Leftrightarrow v·\frac{2v}{g}=5$ . Therefore, $v=\sqrt{\frac{5}{2}g}$ , which gives $v=4.9497$ using $\left|g\right|=9.8$ .

Say the frog starts at the origin $(0, 0)$ at time $t=0$ , and the end of the board starts at the point $(0, 5)$ at time $t=0$ . The frog's initial velocity is given by the vector

$\vec{v_{frog}} = \langle v_x , v_y \rangle$ .

Since there is no acceleration in the x-direction, the frog's velocity in the x-direction at any time will always be equal to

$v_x$ ,

and thus its position along the x-axis will be

$v_x t$ .

There is a constant negative acceleration of

$a = 9.8$

in the y-direction, so the velocity in the y-direction will be

$v_y - a t$

at any time t, and its position along the y-axis will be

$v_y t - \frac{1}{2} a t^2 + x_0$ ,

and plugging in $a = 9.8$ into that equation and noting that $x_0 = 0$ because the frog starts at the origin, the position along the y-axis is

$v_y t - 4.9 t^2$ .

Conservation of momentum tells us that

$M_{frog} v_{frog} + M_{board} v_{board} = 0$ ,

and because the mass of the frog equals the mass of the board, and because the board does not move in the y-direction, we derive that

$v_{board} = - v_x$ .

The point at the end of the board thus has position

$(5 - v_x t , 0)$

at any time t, because it must start at $(5, 0)$ . In order for the frog to make it to the end of the board, its position has to equal the position of the point at the end of the board at some time $t = T$ . Thus we have

$\langle v_x T , - 4.9 T^2 + v_y T \rangle = \langle 5 - v_x T , 0 \rangle$ ,

which means

$v_x T = 5 - v_x T$ and

$- 4.9 T^2 + v_y T = 0$ .

Solving the first equation for T gives

$T = \frac{5}{2 v_x}$ .

Solving the second equation for T gives

$- 4.9 T (T - \frac{v_y}{4.9}) = 0$ ,

$T = {0, \frac{v_y}{4.9}}$ ,

and because at time $t = 0$ the frog is at the origin, we must have

$T = \frac{v_y}{4.9}$ .

Setting the two equations for T equal, we get

$v_y = \frac{(4.9)(5)}{2 v_x}$ ,

$f_1 (v_x , v_y) = v_y - \frac{(4.9)(5)}{2 v_x} = 0$ .

We want to find the minimum value of the function

$f_2 (v_x , v_y) = \sqrt{v_x ^2 + v_y ^2}$

on the curve

$f_1 = 0$ .

A result from multivariable calculus tells us that at the point which minimizes the function along the contour curve of another function, the gradients of the two functions must be parallel, or equivalently, their normal vectors are parallel. This gives us the three equations:

$v_y - \frac{(4.9)(5)}{2 v_x} = 0$ , from above;

$\bigtriangledown f_2 = \lambda \bigtriangledown f_1$ for some real number $\lambda$

$\Longrightarrow \frac{v_x}{\sqrt{v_x ^2 + v_y ^2}} = \lambda \frac{(4.9)(5)}{2 v_x ^2}$ , and

$\frac{v_y}{\sqrt{v_x ^2 + v_y ^2}} = \lambda$ .

Multiply the factor $\sqrt{v_x ^2 + v_y ^2}$ out of the denominator of the last two equations to get:

$v_x ^3 = \lambda \frac{(4.9)(5)}{2} \sqrt{v_x ^2 + v_y ^2}$ and

$v_y = \lambda \sqrt{v_x ^2 + v_y ^2}$ .

Substituting $v_y$ into the first equation, we get

$v_x ^ 3 = v_y \frac{(4.9)(5)}{2}$ ,

and the relation $v_y = \frac{(4.9)(5)}{2 v_x}$ gives us

$v_x ^ 4 = (\frac{(4.9)(5)}{2})^2$ ,

and because $v_x$ must be a real number, and since the frog must be moving along the x-axis in the positive sense, $v_x$ must be the positive real solution to this equation,

$v_x = \sqrt{\frac{(4.9)(5)}{2}} = 3.5$ .

$v_y = \frac{(4.9)(5)}{2 v_x} = 3.5$ ,

and the speed will be the value

$S = \sqrt{3.5 ^2 + 3.5 ^2} = 3.5 \sqrt{2} \approx 4.950$ .

Solution with simple maths: Let the horizontal velocity be vx, the vertical velocity vz and the total velocity v. The time the vertical movement needs is 2vz/9.8 The time the horizontal movement needs is 2.5/vx (The frog only travels 2.5m relative to the ground because the board comes towards him with the same velocity vx) These times must be of same length -> 2.5/vx=2vz/9.8 Also: vx^2=v^2-vz^2 Squaring the first equation, substituting vx^2 with v^2-vz^2 and solving for v gives: v=✓[(150+vz^4)/vz^2] If you plot this (use x for vz) you will see that the minimal y-value is 4,95.

Using momentum conservation it is easy to see the velocity the frog jumps off is same as the velocity the board goes in the other direction.

So,to reach the other side the frog need only traverse $\frac{1}{2}$ the total length= $2.5 m$

knowing the highest projectile range to be $\frac{v^2}{g}$ it is a matter of calculation to find the answer to be

$\boxed{4.94}$ approximately.

The actual correct answer is epsilon (namely as close to 0 as you wish). Aim almost straight up at near infinite velocity. The board slides out from underneath while the frog has almost 0 lateral velocity. Some (long) time later, the frog lands at the end of the board. To reach the conclusion of the 45 degree takeoff, you need to minimize SPEED, not ground speed.

Also, FYI, the board mass is irrelevant. Consider the problem in the frame of reference of the center of mass of the two bodies. In this frame the frog minimizes its take off speed by aiming at 45 degrees.