Particle Flying Through A Ramp

The forces acting on the body is given in the free body diagram.

${ f }_{ k }$ represents the kinetic frictional force acting on the body.

Now, The net acceleration of the body,

$a=-[mg\sin { { 40 }^{ 0 } } +{ \mu }_{ k }N]$ , where ${ \mu }_{ k }$ is the coefficient of kinetic friction.

Let $v$ be the velocity of particle at B and $u$ be the velocity of particle at A ,

Now, using Newton's equations,

${ v }^{ 2 }={ u }^{ 2 }+2as$ , where $s$ is the distance between A and B .

$\Rightarrow { v }^{ 2 }={ u }^{ 2 }-2[mg\sin { { 40 }^{ 0 } } +{ \mu }_{ k }N]s$ Substituting values,

$\Rightarrow { v }^{ 2 }={ 20 }^{ 2 }-2.[1\times 10\sin { { 40 }^{ 0 } } +0.8\times 1\times 10\times \cos { { 40 }^{ 0 } } ].\frac { 4 }{ \cos { { 40 }^{ 0 } } }=268.872029 m/s$

{Note: $N=mg\cos { { 40 }^{ 0 } }$ }

At B , the body leaves the ramp and undergo projectile motion. The body will be projected at an angle ${ 40 }^{ 0 }$ with the horizontal.

Since both both ramps are at same level, the projectile lands perfectly on other ramp when the range of projectile is maximum.

Now we could use the equation for range of projectile,

$R=\frac { { { \left( { u }_{ 1 } \right) } }^{ 2 }sin2\theta }{ g }$ , where

${ u }_{ 1 }\longrightarrow$ Velocity with which the projectile is projected

$\theta \longrightarrow$ Angle made by projectile with horizontal at time of projection.

Now we have ${ { \left( { u }_{ 1 } \right) } }^{ 2 }$ = ${ v }^{ 2 }$ = $268.872029m/s$ and $\theta ={40}^{0}$

$\therefore \quad x=R=\frac { { { 268.872029\times } }sin{ 80 }^{ 0 } }{ 10 } =\boxed{26.4787m}$

PS: If you have doubt about how the equation of "Range of projectile" came through, let me know. I'll give the explanation.

First, find the deceleration as the particle moves along the ramp.. $\text{ Flim }= uR = 0.8\cdot 10\cdot \cos(40^\circ) = 8\cos40^\circ$
Horizontal Component of weight is $10\sin(40^\circ)$
Therefore..
$a=-\dfrac{8\cos40^\circ+10\sin40^\circ}{1}= -(8\cos40^\circ+10\sin40^\circ)$

From here we can simply use SUVAT equations..
$u=20\text{ m/s}, \quad v=\, ? , \quad s=\frac{4}{\cos40^\circ}, \quad a=-(8\cos40^\circ+10\sin40^\circ)$

v=\sqrt{20^{2}+(2\cdot-(8\cos40^\circ+10\sin40^\circ)\cdot \frac{4}{\cos40^\circ}

$v=16.40\text{ m/s}$ (by the top of the ramp)

Now it's a normal projectile question:
$u=16.40\sin40^\circ, \quad v=0, \quad t=\, ? , \quad a=-10$

$t= \frac{-16.40\sin40^\circ}{-10} = 1.054$ seconds for half the time
Therefore total time is $1.054\times 2 = 2.108$

$s=x=ut$
$x = 16.40\cos(40^\circ) \cdot 2.108$
$x = 26.47$