Sam (75 kg) takes off up a 50-m-high, 10­ frictionless slope on his jet-powered skis. The skis have a thrust of 200 N. He keeps his skis tilted at 10­^o after becoming airborne, as shown in FIGURE CP8.66. How far does Sam land from the base of the cliff?

Mass = 75 kg
Takes off at 50 m height
Angle of projection 10^o
The skis have a thrust of 200 N.
Ski is tilted at 10^o after becoming airborne.

We know that work done by a constant force F in moving a body a distance d is given by

W=F d.

Potential energy of a mass at height h is given as mgh.

KE of an object moving with velocity v is given by 1/2mv²

The displacement S of an object moving with acceleration and initial speed u is given by

$s=ut+\frac{1}{2}a{t}^2$

From the law of conservation of energy

Kinetic energy at the release point is equal to work done minus potential energy

$$\begin{gathered} Fd-mgh=\frac{1}{2}m{v}^2 \\ m{v}^2=2(Fd-mgh) \\ {v}^2=\frac{2(Fd-mgh)}{m} \\ v=\sqrt{\frac{2(Fd-mgh)}{m}} \end{gathered}$$

Substitute the given values and calculate v

$$\begin{gathered} v=\sqrt{\frac{2\left(\right(200N)\times (50mcsc\left(\theta \right))-(75kg)\times (9.81m/s2)\times (50m\left)\right)}{75kg}} \\ v=16.65m/s \end{gathered}$$

Now find time it takes for Sam to touch down

$$\begin{gathered} -50m=(16.6m/s)sin\left(\theta \right)t-\frac{1}{2}(9.8m/s^2)t^2 \\ \mathrm{simplify} \\ 4.905t^2-2.891t-50=0 \end{gathered}$$

Solve the quadratic equation

we get

t= - 2.912 s, and 3.501 s

Drop the negative value so t=3.501 sec.

So total distance will be horizontal component of velocity multiplied by time

$d=v\cos \left(\theta \right)t=(16.65m/s)\cos (10°)\times 3.501s=57.41m$