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In fig. a rescue plane flies at 198 km/h ( = 55.0 m/s)and constant height h = 500 m toward a point directly over a victim , where a rescue capsule is to land . (a) what should be the angle phi of the pilots line of sight to the victim when the capsulerelease is made ?(b) as the capsule reaches the water , what is the velocity vecv? |
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Answer» Solution :One released , the capsule is projectile , so the horizontal and verical MOTIONS can be considered separately (we need not consider the actual curved path of the capsule) In fig. we see that `PHI` is given by `phi = tan^(-1) x/h`  where x is the horizontal coordinate of the victim (and of the capsule when it hits the water ) and h= 500 m . we should be able to find x with eq. ` x - x_0 (v_ cos theta_0) y` To find t, we next consider the vertical motion and specifically eq. ` x- y_0 = (v_0 sin theta_0) t - 1/2 "gt"^2` Here verical displacement y - `y_0` of the capsule is -500 m( the negative value indicates that the capsule MOVES downward). so ` - 500m = (55.0 m//s) sin 0^@) y - 1/2 (9.8 m//s^2) t^2` solving for t . we find t = 10.1 s. using that valuein eq. yields `x - 0 = (55.0 m//s) (cos 0^@) (10.1 s)` x = 555.5 m The eq. gives us `phi = tan^(-1) (555.5m)/(500 m) = 48.0^@` (b) when the capsule reaches the water ` v_x = v_0 cos theta_0 = (55.0 m//s) (cos 0^@) = 55.0 m//s` using eq. and the capsule.s time of fall t = 10.1 s , we also find that when the capsule reaches the water , `v_y = v_0 sin theta_0 - "gt"` `= (55.0 m//s) (sin 0^@)- (9.8 m//s^2)(10.1 s)` = -99.0 m/s Thus , at the water `VECV = (55.0 m//s)hati - (99.0 m//s)HATJ ` From eq. the magnitude and angle of `vec v` are `v = 113 m//s andtheta = - 60.9^@` |
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