A 55-kg skier starts from rest at the top of a ski jump, point A in Fig. 6–48, and travels down the ramp. If friction and air resistance can be neglected, (a) determine her speed when she reaches the horizontal end of the ramp at B. (b) Determine the distance s to where she strikes the ground at C.

The law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another. The total energy of a closed system remains constant.

The mass of the skier is\(m = 55\;{\rm{kg}}\).

The steepness of the ramp is\(\theta = {30^{\rm{o}}}\).

The distance between the highest point of the ramp and the ground is\(d = 45\;{\rm{m}}\).

The distance between the horizontal end of the ramp and the ground is\(d' = 4.4\;{\rm{m}}\)

The distance covered by the skier on the ramp is:

\(\begin{aligned}{h_{\rm{A}}} &= d - d'\\ &= 45\;{\rm{m}} - 4.4\;{\rm{m}}\\ &= 40.6\;{\rm{m}}\end{aligned}\)

The speed of the skier at the top of the ramp is \({v_{\rm{A}}} = 0\), and the speed of the skier at the end of the ramp is \({v_{\rm{B}}}\). Let point B be the zero location of potential energy\(\left( {{h_{\rm{B}}} = 0} \right)\). Applying the law of conservation of energy at the top (point A) and bottom (point B), you get:

\(\begin{aligned}K{E_{\rm{A}}} + P{E_{\rm{A}}} &= K{E_{\rm{B}}} + P{E_{\rm{B}}}\\\frac{1}{2}mv_{\rm{A}}^2 + mg{h_{\rm{A}}} &= \frac{1}{2}mv_{\rm{B}}^2 + mg{h_{\rm{B}}}\\0 + mg{h_{\rm{A}}} &= \frac{1}{2}mv_{\rm{B}}^2 + 0\\mg{h_{\rm{A}}} &= \frac{1}{2}mv_{\rm{B}}^2\end{aligned}\) … (i)

Equation (i) implies that the potential energy at the top is equal to the kinetic energy at the bottom. Rearrange equation (i) for the speed of the skier at the bottom.

\(\begin{aligned}{v_{\rm{B}}} &= \sqrt {2g{h_{\rm{A}}}} \\ &= \sqrt {2\left( {{\rm{9}}{\rm{.8}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {40.6\;{\rm{m}}} \right)} \\ &= 28.2\;{\rm{m/s}}\end{aligned}\)

Projectile motion is the path traveled by an object launched in the air and allowed to move under the influence of gravity. For example, kicking a football, firing a cannonball, firing a bullet from the gun, etc.

The slope of the ramp is given by:

\(\begin{aligned}{y_{{\rm{slope}}}} &= - x\tan \theta \\ &= - x\tan {30^{\rm{o}}}\end{aligned}\)

The initial velocity of the skier along the y-axis is 0, and its velocity along the x-axis is\({v_{\rm{B}}}\). The motion of the skier along the positive vertical direction is given by:

\(\begin{aligned}x &= {v_{\rm{B}}}t\\t &= \frac{x}{{{v_{\rm{B}}}}}\end{aligned}\)

Using the following kinematic equation of motion, you get:

\(\begin{aligned}{y_{{\rm{projectile}}}} &= {y_0} - \frac{1}{2}g{t^2}\\ &= {y_0} - \frac{1}{2}g{\left( {\frac{x}{{{v_{\rm{B}}}}}} \right)^2}\end{aligned}\) … (ii)

The skier lands on the ground at the intersection of the two paths such that\({y_{{\rm{slope}}}} = {y_{{\rm{projectile}}}}\).

\(\begin{aligned} - x\tan {30^{\rm{o}}} &= {y_0} - \frac{1}{2}g{\left( {\frac{x}{{{v_{\rm{B}}}}}} \right)^2}\\g{x^2} - x\left( {2v_{\rm{B}}^2\tan {{30}^{\rm{o}}}} \right) - 2{y_0}v_{\rm{B}}^2 &= 0\end{aligned}\)

Rearrange the following equation to solve for x.

\(\begin{aligned}x &= \frac{{\left( {2v_{\rm{B}}^2\tan {{30}^{\rm{o}}}} \right) \pm \sqrt {{{\left( {2v_{\rm{B}}^2\tan {{30}^{\rm{o}}}} \right)}^2} + 8g{y_0}v_{\rm{B}}^2} }}{{2g}}\\ &= \frac{{\left( {v_{\rm{B}}^2\tan {{30}^{\rm{o}}}} \right) \pm \sqrt {{{\left( {v_{\rm{B}}^2\tan {{30}^{\rm{o}}}} \right)}^2} + 2g{y_0}v_{\rm{B}}^2} }}{g}\\ &= - 7.09\;{\rm{m,}}\;{\rm{100}}{\rm{.8}}\;{\rm{m}}\end{aligned}\)

Neglecting the negative value of the distance, you get:

\(x = 100.8\;{\rm{m}}\)

Therefore,

\(\begin{aligned}x &= s\cos {30^{\rm{o}}}\\s &= \frac{x}{{\cos {{30}^{\rm{o}}}}}\\ &= \frac{{100.8\;{\rm{m}}}}{{\frac{{\sqrt 3 }}{2}}}\\ &= 116.39\;{\rm{m}}\end{aligned}\)

Thus, the distance s is 116.39 m.