Calculate the distance d she travels along the incline before landing.

Question

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 29 m/s as in the figure. The landing incline below her falls off with a slope of θ = 58◦ . The acceleration of gravity is 9.8 m/s 2 .

Answer 1

The solution to this question is straight forward. Imagine a right triangle in which the hypotenuse is the length along the incline, the vertical side is a measure of the vertical distance travelled by the jumper (1/2(98)*t²= 4.9t²) and the horizontal side is the distance covered by the jumper (29t).

If the angle between the horizontal and the slope is θ, then;

tan(θ) = 4.9t2/29t = 4.9t/29

so

t = 29[tan(θ)]/(4.9)

If we set θ = 58 we get t= 9.47 sec,

t = 29[tan(θ)]/(4.9) = 29(1.6)/(4.9) = 9.47 sec

and the distance along the slope is

d = 29(9.47)/[cos(58)] = 274.63/(0.53) = 517.3 meters