Question

A ski jumper travels down a slope and
leaves the ski track moving in the horizon-
tal direction with a speed of 30 m/s as in the
figure. The landing incline below her falls off
with a slope of = 48º.
The acceleration of gravity is 9.8 m/s².
30 m/s

Answer 1

Answer:

304.86 metres

Explanation:

The x and y cordinates are $dcos\theta$ and $dsin\theta$ respectively

The horizontal distance travelled, $x=v_{ox}t=dcos\theta$

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

Since $y=0.5gt^2=dsin\theta$, we substitute t with the above and obtain

$0.5g(\frac{dcos\theta}{v_{ox}})^2=dsin\theta$

Making d the subject we obtain

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

$d=\frac{2*30^2sin48}{9.8cos^248}$

d=304.8584

d=304.86m