A snowboarder glides down a 50-m-long, 15° hill. She then glides horizontally for 10 m before reaching a 25° upward slope. Assume the snow is frictionless.

a. What is her velocity at the bottom of the hill?

b. How far can she travel up the 25° slope?

The snowboarder slides on hill for distance $H=50m$
The hill is inclines at the angle $15°$
Horizontal distance covered by snowboarder after passing hill, $S=10m$
After crossing horizontal distance, the inclined upward slope is$25°$

At point A, the initial velocity of the snowboarder is $u=0m/sec$
Let the final velocity is represented by $v$
The acceleration due to gravity is $g=10m/s^2$
The acceleration experienced by the snowboarder at the inclination is $a=g\sin (15°)$
$$\begin{gathered} \Rightarrow a=10\sin (15°) \\ a=2.58m/s^2 \end{gathered}$$

Now, using the equation of motion,
$v^2-u^2=2aH$
Substitute the unknown values, to find out $v$

localid="1648208779028" $$\begin{gathered} v^2-0=2\left(2.58\right)\left(50\right) \\ {v}^2=258 \\ \Rightarrow v=16m/s \end{gathered}$$

Thus, the velocity of the snowboarder at the bottom of the hill is 16m/sec.

After coming to the bottom of the hill, the snowboarder travelled the distance of 10m horizontally. Her speed remains same before reaching the upward slope as shown in the figure.
Now, at the maximum distance she could travel up on the hill $S$,

The final velocity would be $v_f=0m/sec$

The initial velocity is already known, i.e. role="math" localid="1648209252088" $v_i=16m/sec$
The acceleration experiences by her is

role="math" localid="1648209367859" $$\begin{gathered} a=g\sin \left({25}^0\right) \\ a=10(0.42) \\ \Rightarrow a=4.2m/s^2 \end{gathered}$$

Write the equation of motion to determine the maximum distance she could travel,

${v_f}^2-{v_i}^2=2aS$
Substitute the known values,

role="math" localid="1648209527195" $$\begin{gathered} 0-{16}^2=2(-4.2)S \\ S=\frac{256}{8.4} \\ \Rightarrow S=30.4m \end{gathered}$$
Thus, snowboarder could travel to maximum 30.4 m on the 25^oslope.