Question #6c16b

1 Answer
Jan 6, 2018

v_"xy" = 243 m/s

Bomb will horizontally travel 183.5 m

Explanation:

Data
v_x = 15 m/s

v_y = (2gh)^(1/2) = (2(9.8)(3000))^(1/2) = 242.5 m/s

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

v_"xy" = ((15)^2 + (242.5)^2)^(1/2) = color(red)242.95 m/s

theta = tan^(-1) (242.95/15) = 86.5^o

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

x = 3000/(tan (86.5))

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= (15m/s) 20 s = 300 meters which is longer than 183.5 meters.

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

t = x/v_x =183.5/15 = 12.3 seconds