Answer to Question #251265 in Mechanics | Relativity for john

Question

a sky jumper speeds down a slope and exits the ski track in horizontal direction at a velocity of 25 m per second where the landing inclined falls of the slope of 33 °

solve the following

what is the halftime of flight of the sky jumper

what is the range of the sky jumper when it lands

Answer 1

Determine the height between the point of jump and point of landing:

$$h=\frac{gt^2}2.$$

Determine the range:

$$R=vt.$$

The relation of R to h is

$$h=R\tan\theta=vt\tan\theta.\\\space\\ vt\tan\theta=\frac{gt^2}2,\\\space\\ t=\frac{2v\tan\theta}{g}=3.31\text{ s},$$

the halftime is 1.66 s, the range is

$$R=25·3.31=82.75\text{ m}.$$