Question #6c16b

Question

893 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-06T21:05:30

#v_"xy" = 243 m/s#

Bomb will horizontally travel #183.5# m

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