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

304.86 metres

Step-by-step 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=(dcos\theta)/(v_(ox))`

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

`0.5g((dcos\theta)/(v_(ox)))^2=dsin\theta`

Making d the subject we obtain

`d=(2v_(ox)^2sin\theta)/(gcos^2\theta)`

`d=(2*30^2sin48)/(9.8cos^248)`

d=304.8584

d=304.86m