Answer to Question #106144 in Trigonometry for simran

Question

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?

Answer 1

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 45^o 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

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

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

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

$2x=240+300$

$x=270m$