91Ó°ÊÓ

The skier is moving at the speed of $u=3m/sec$

The skier starts moving along the horizontal plane inclines at 10^o

The final velocity $v=15m/sec$

Let the acceleration due to gravity is localid="1648194296435" $g=9.8m/s^2$

The skier would be experiencing the acceleration given by $a=g\sin \left(\mathrm{\AE Ÿ}\right)$

Here, $\mathrm{\AE Ÿ}$is the angle of inclination of the horizontal plane.

Let the distance traveled by the skier on the inclined surface is S . The acceleration experienced by the skier is
$$\begin{gathered} a=(9.8)\sin (10°) \\ a=1.7m/s^2 \end{gathered}$$
Using the equation of motion,
$$\begin{gathered} v^2-u^2=2aS \\ {\left(15\right)}^2-{\left(3\right)}^2=2(1.7)S \\ 225-9=3.4S \\ S=\frac{216}{3.4}=63.5m \end{gathered}$$
Thus, the distance covered by the skier on the inclined horizontal plane is 63.5m

Let the time taken by the skier to travel the distance of 63.5 m is $t$
Using the equation of motion,
$$\begin{gathered} v=u+at \\ 15=3+(1.73)t \\ t=\frac{12}{1.73}=6.9sec \end{gathered}$$

Thus, the time taken by the skier to reach the bottom point is 6.9 sec