Physics, Man skis up a slope.

I interpret the question as: "How far does he go before he stops?"

In this case the appropriate equation is v² = u² + 2as  where v is final velocity (0), u is initial velocity (8.1m/sec), a is acceleration and s is distance.

The acceleration here is -9.8*sin(20°) m/sec², the component of gravitational acceleration acting parallel to the slope (negative because it is retarding the skier).

0 = 8.1² - 2*9.8*sin(20°)*s

s = 8.1²/(2*9.8*sin(20°)) metres

$${\mathtt{s}} = {\frac{{{\mathtt{8.1}}}^{{\mathtt{2}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{20}}^\circ\right)}\right)}} \Rightarrow {\mathtt{s}} = {\mathtt{9.787\: \!286\: \!055\: \!842\: \!569\: \!1}}$$

s ≈ 9.8 metres

I'm far more into math, then I am into physics, but I'll take a shot at this:

the formula for location is:  d  = -4.9t² + vt

If the velocity is  8.1  along the slope, the vertical component of this will be found by using sin:

    sin(20°)  =  v/8.1   --->   v = 2.770

The formula becomes:  d  = -4.9t² + 2.77t

Since this formula describes a parabola, the maximum point (the stopage point) is halfway between where the distance is 0 (when it starts) and where it again becomes 0.

Solve:  0  = -4.9t² + 2.77t     --->     0  =  t(-4.9t + 2.77)     --->  t = 0  and  t =  0.5653

Halfway between 0  and  0.5653  is  0.28 seconds.

Again, no guarantee.