Question

a skim jumper travels downa a slope and leaves the ski track moving in the horizontal direction with a speed of 20 m/s as in the figure. The landing incline belowe her falls off with a slope of 39degrees. the acceleration of gravity is 9.8m/s2 . calculate the distance d she travels along the incline before landing

Answer 1

Answer:

85.1 m

Explanation:

The x and y coordinates are as follows

$x=dcos\theta$

$y=dsin\theta$ where d is distance travelled and $\theta$ is the slope of landing

The horizontal distance travelled is $x=vt= dcos\theta$ where t is time take to travel and v is the velocity

Making t the subject, $t=\frac {dcos\theta}{v}$

Also, the vertical distance $y=0.5gt^{2}=dsin\theta but t= t=\frac {dcos\theta}{v}$

Therefore,

$dsin\theta=0.5g(\frac {dcos\theta}{v})^{2}$ and making d the subject of the formula

$d=\frac {2v^{2}sin\theta}{gcos^{2}\theta}$

Substituting v for 20 m/s, $\theta$ for $39*{o}$ and g for $9.81 m/s^{2}$

$d=\frac {2*(20 m/s)^{2}*sin39^{o}}{9.8 m/s ^{2}cos^{2}39^{o}}$

d=85.06100825

Rounding off, d=85.1 m