91Ó°ÊÓ

We know that the angle of inclination is $\mathrm{θ}=20°$, height of the inclination is $h=2.0\mathrm{m}$, the coefficient of friction between the bike tire and the inclination surface is $\mathrm{μ}=0.02$, width of the pool is $d=10.0\mathrm{m}$, acceleration due to gravity $g=9.81\mathrm{m}/{\mathrm{s}}^2$ and the initial speed of the bike is $v_{\mathrm{mtight}}>\mathrm{i}=11.0\mathrm{m}/\mathrm{s}$. We have to calculate the distance of landing position from the end of inclination.

Let's consider the information given in the question:

The equation force motion is given as

$ma=mg\sin \mathrm{θ}+F_f$

Where,

$F_f$is the friction force

$ma=mg\sin \mathrm{θ}+\mathrm{μ}mg\cos \mathrm{θ}$

$ma=mg(\sin \mathrm{θ}+\mathrm{μ}\cos \mathrm{θ}$

$⇒a=g(\sin \mathrm{θ}+\mathrm{μ}\cos \mathrm{θ})$

Substitute the values

$a=(9.8m/s^2)(\sin \mathrm{θ}{20}^0+(0.02\left)\cos {20}^0\right)$

$a=(9.8m/s^2)(0.36)$

$a=3.54m/s^2$

Therefore, the length of the ramp is given as,

$\sin \mathrm{θ}=\frac{h}{L}$

Where,

$h$is the height$=2m$

$L$is the length of the ramp

So,

$L=\frac{h}{Sin\mathrm{θ}}$

$⇒L=\frac{2m}{Sin{20}^0}$

$L=5.85m$

Let's consider the equation of motion:

$v^2=u^2+2(-a)L$

where,

$v$is the initial velocity

$u$is the instant velocity

$v^2=u^2-2al$

role="math" localid="1648361146542" $v^2={(11m/s)}^2-2(3.54m/s^2)(5.85m)$

role="math" localid="1648361542478" $v^2=79.58m^2/s^2$

role="math" localid="1648361565005" $⇒v=\sqrt{79.58m^2/s^2}$

role="math" localid="1648361593999" $v=8.92m/s$

The length of the pool is $10m$.

Calculate the initial velocity has two components along the horizontal direction and the vertical direction as given as,

$v_x=v\cos {20}^0$

$v_x=(8.92m/s)\cos {20}^0$

$v_x=8.38m/s$

Compute for $v_y$

$v_y=v\sin {20}^0$

$v_y=(8.92m/s)\sin {20}^0$

$v_y=3.05$

Therefore, the equation along the vertical direction is given as:

$y=y_0+u_{y}t+\frac{1}{2}{a}_y{t}^2$

$0=2\mathrm{m}+(3.05\mathrm{m}/\mathrm{s})t−\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t^2$

role="math" localid="1648363560106" $⇒−\left(4.9\mathrm{m}/{\mathrm{s}}^2\right)t^2+(3.05\mathrm{m}/\mathrm{s})t+2\mathrm{m}=0$

On solving we get,

$t=1.02$

The position of the motorcycle during the projectile time is given as,

$x=u_{x}t$

$x=(8.38\mathrm{m}/\mathrm{s})(1.02\mathrm{s})$

$x=8.547$

Here we see that the distance or position of the motorcycle is less than the length of the pool. So, the motorcycle cannot cross the pool and he falls into the pool.

The distance or position of the motorcycle is 8.54m and it is less than the length of the pool. So, the motorcycle cannot cross the pool and he falls into the pool