Answer to Question #96265 in Mechanics | Relativity for Sarah

Question

A ski jumper travels down a slope and leaves
the ski track moving in the horizontal direction with a speed of 23 m/s as in the figure.
The landing incline below her falls off with a
slope of θ = 56◦.The acceleration of gravity is 9.8 m/s2.
Don't round answer

Answer 1

Let ski jumper falls at a point which is $y$ $m$ below the initial point and $x\ m$ far from the base.

Then

Slope = $\frac{y}{x}=\tan56\degree=1.4826$

So,

$y=1.4826x\ \ \ \ .......(1)$

From equation of motion

$x=23t\ \ \ \ .....(2) \\where \ t=time\ of\ flight$

and ,

$y=4.9t^2\ \ \ \ .....(3)$

Putting the value of y and x obtained from (2) and (3) into (1)

The,

$4.9t^2=1.4826×23t$

Or

$t=\frac{1.4826×23}{4.9}=7 \ sec$

And,

$x=23t=161\ m$

And,

$y=4.9t^2=4.9×49=240.1\ m$