Question #96265

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

Expert's answer

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,@$y=4.9t^2=4.9×49=240.1\ m@$

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