Question #6c16b

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Answer 1 · 2018-01-06T21:05:30

$v_{xy} = 243 \frac{m}{s}$ $v_"xy" = 243 m/s$

Bomb will horizontally travel $183.5$ $183.5$ m

Data
$v_{x} = 15 \frac{m}{s}$ $v_x = 15 m/s$

$v_{y} = {(2 g h)}^{\frac{1}{2}} = {(2 (9.8) (3000))}^{\frac{1}{2}} = 242.5 \frac{m}{s}$ $v_y = (2gh)^(1/2) = (2(9.8)(3000))^(1/2) = 242.5 m/s$

Calculating bomb speed and bottom-angle $θ$ $theta$ using above speeds (vectors):

$v_{xy} = {({(15)}^{2} + {(242.5)}^{2})}^{\frac{1}{2}}$ $v_"xy" = ((15)^2 + (242.5)^2)^(1/2)$ = $242.95$ $color(red)242.95$ m/s

$θ = {tan}^{- 1} (\frac{242.95}{15}) = {86.5}^{o}$ $theta = tan^(-1) (242.95/15) = 86.5^o$

Calculating bomb location "x" using angle $θ$ $theta$ and altitude "y":

$x = \frac{3000}{tan (86.5)}$ $x = 3000/(tan (86.5))$

$x = 183.5$ $x = color(red)183.5$ meters

Double-checking if bomb will be in the "air" or in the "ground"
Using horizontal-speed

$x = (v_{x}) t = (15 \frac{m}{s}) 20 s = 300$ $x = (v_x) t= (15m/s) 20 s = 300$ meters which is longer than $183.5$ $183.5$ meters.

So, bomb will be at ground after 20 seconds at distance of 183.5 meters.
Bomb dropped only in:

$t = \frac{x}{v_{x}} = \frac{183.5}{15} = 12.3$ $t = x/v_x =183.5/15 = 12.3$ seconds