Question

A $60\mathrm{kg}$skier starts from rest at height$\text{H=20m}$above the end of a ski-jump ramp (Figure) and leaves the ramp at angle$\mathrm{\theta }=28°$. Neglect the effects of air resistance and assume the ramp is frictionless.

1. What is the maximum height hof his jump above the end of the ramp?
2. If he increased his weight by putting on a backpack, would hthen be greater, less, or the same?

Short answer

1. The maximum height h of the jump above the end of ramp is 4.404 m.
2. The effect of weight on that height, h is zero.

Step-by-step solution

Step 1: Given

1. A mass of skiers$m=60\mathrm{kg}$
2. Starting height of skier,role="math" localid="1663125898068" $\text{H=20m}$
3. The angle with which skiers leave the ramp is$\theta ={28}^o$

Determining the concept

Using the energy conservation law, find the velocity with which the skier leaves the ramp, and by using the given angle and kinematic equation, find the height of the skier after he leaves the ramp. According to the law of energy conservation, energy can neither be created nor destroyed.

The formula is as follows:

$$\begin{gathered} v=v_0^2+2as \\ Energy=PE+KE \\ =mgh+\frac{1}{2}m{v}^2 \end{gathered}$$

where, KE is kinetic energy, PEis potential energy, m is mass, v is velocity, g is an acceleration due to gravity, a is an acceleration, s is displacement and h is height.

(a) Determining the maximum height h of the jump above the end of ramp 

The skier starts from a height of 20 m at rest. So, there is only potential energy, and when the skier is at the bottom of the ramp, he has only kinetic energy. So,

$$\begin{gathered} \text{mgH=}\frac{1}{2}m{v}^2 \\ v=\sqrt{2gH} \\ v=19.79\text{m/s} \end{gathered}$$

At maximum height, the speed of objects is zero. From the diagram below, resolve the initial velocity after leaving the ramp along the x and y axis.

From the 2^nd kinematic equation,

$v^2=V^2+2as$

where $a$is acceleration and $s$is vertical distance.

$$\begin{gathered} 0=V\sin {\left(\theta \right)}^2-2\left(9.8\right)h \\ h=\frac{V\sin {\left(\theta \right)}^2}{2\times 9.8} \\ h=\frac{{\left(19.79\sin \left(28\right)\right)}^2}{2\times 9.8} \\ h=4.\text{404m} \end{gathered}$$

Hence, the maximum height h of the jump above the end of the ramp is $44.04m$

(b) Determining the effect of weight on that height

From the derived equation of height,

$h=\frac{v-\sin {\left(\theta \right)}^2}{2\times 98}$

It can be predicted that there will be no effect of weight ontheheight as the equation is independent of mass.

Hence, the effect of weight on that height h is zero.