Question

A shell when projected at an angle of
tan-1 1/3
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

This particular question is a combined case of physics and trigonometry ,where shell is being projected at an angle,means we have to apply the concept of projectile motion of an object.

the diagrammatic presentation of the problem is attached as an image.

Let O be the point of projection.

Let the shell be projected with an initial velocity of $u$ m/sec at an angle $tan^{-1}\frac{1}{3}=\alpha(let)$ and it reached point A ,60 m short of the target point T.So in this case range of the projectile motion is OA.

Secondly when the shell is fired with an initial velocity of $u$ m/sec at an angle of 45$\degree$ ,it reached point B, 80 m beyond the target point T.

So in this case range of projectile motion is OB.

Range in projectile motion is given by

$$R=\frac{u^2sin2\alpha}{g}$$

where $g$ is an acceleration due to gravity.

CASE I : In this case $(tan^{-1}\frac{1}{3}=\alpha \implies tan\alpha=\frac{1}{3})$

$OA = \frac{u^2sin2\alpha}{g} \\Now \space sin2\alpha = \frac{2tan\alpha}{1+tan^2\alpha} \space ................(1) \\tan\alpha=\frac{1}{3}\\putting \space the \space value\space of \space tan\alpha=\frac{1}{3} \space in (1)\\we \space get \space sin2\alpha=\frac{2*\frac{1}{3}}{1+\frac{1}{9}}=\frac{3}{5}.\\\\OA=\frac{u^2*\frac{3}{5}}{g}\\OA=\frac{3u^2}{5g}$

similarly

CASE II: In this case

$OB=\frac{u^2sin(2*45)\degree}{g}\\OB=\frac{u^2sin90\degree}{g}=\frac{u^2}{g} \space (\because$ $sin90\degree=1)$

$Now \\OA=\frac{3}{5}*\frac{u^2}{g}\\OA=\frac{3}{5}*OB \space \space (\because\space OB=\frac{u^2}{g})\\Now \space let\space OA=x \space meters.\\\therefore x=\frac{3}{5}*(x+140)\\or, x=\frac{3x+420}{5}\\or, 5x=3x+420\\or,5x-3x=420\\or,2x=420\\or, x=\frac{420}{2}=210\space meters.\\\therefore OA=210\space meters.$

Now from the above diagram we can see target is at a distance of OT meters from the point of projection.

$\therefore OT=OA+AT\\or, OT=210+60=270\space meters.$

ANSWER: The target is 270 meters away from the point of projection.