In James bond movie ''The Spy Who Loved Me'', James skied down a snowy slope in an attempt to escape from the killer. Unfortunately, the killer had a higher skiing speed than James's, they are `40m//s` and `40m//s` respectively. Now consider an alternative version of the movie.
James noted that he and the killer carried the same kind of rifle' and estimated that their masses M (including body mass, skis, weapon, backpack, ect) were about the same. Recalling high school physics knowledge, James realised that each time he fired his weapon back at the killer, his momentum would change, whereas when the killer fired, the killer's momentum would also change. Evergy time James fired a bullet, the killer would fire back accordingly. Assume that all bullets missed their targets (otherwise this exercise would terminte). Given `M = 100 kg`, mass and the muzzle velocity of a bullet are `m = 0.02 kg` and `v = 500 m//s` respectively

Traveling with the final velocity obtained in part above question, James Bond escaped by sliding down a cliff at an inclination of `20º`. After 20 second, he opened his parachute and landed safely. The average air drag during his fall is `600 N` in both the vertical and horizontal directions, as long as the velocity components are nonzero. Calculate the horizontal landing position of James Bond from the of edge the cliff. (Neglect the distance he traveled with a parchute.) `cos `20º = 0.94`

Options

A. `133m`
B. `233m`
C. `33m`
D. `333m`

Answer

A

Explanation

James final velocity `v_(f)`
`v_(f) = 40 + 25 xx (92)/(100) = 42.3`
`:. V_(x_(0)) = v_(f) cos 20 = 40 m//s`
Retardation, `a = (-600)/(1000) = -6`
`v_(x) = 0 rArr 40 - 6t = 0`
Final horizontal component `t = (20)/(3)`
`x = 40 xx (20)/(3) - (1)/(2) xx 6 xx ((20)/(3))^(2) = 133 m`