Answer to Question #106144 in Trigonometry for simran

Question #106144
A shell when projected at an angle of
3
1
tan −1
to the horizon falls 60 m. short of the target. When
it is fixed at an angle of o
45 to the horizon, it falls 80m beyond the target. How far is the target
from the point of projection?
1
Expert's answer
2020-03-24T13:09:30-0400

Range of a projectile is given by "\\dfrac{v^2\\sin2\\theta}{g}"


Let the target from the point of projection be 'x'

when "\\theta" is "tan^{-}3" it is said that the shell fells 60 m short of target ,

then range is 'x-60'


when "\\theta" is 45o then it is said that the shell 80 m beyond the target

then range is 'x+80'

if we divide both the ranges then we will get

"\\frac{Range_1}{Range _2}=\\frac{\\sin2\\theta_1}{\\sin2\\theta_2}"


"\\dfrac{x-60}{x+80}=\\dfrac{3\/5}{1}"


"5(x-60)=3(x+80)"

"2x=240+300"

"x=270m"



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