Question

A projectile aimed at a mark which is in the horizontal plane through the point of projection falls a cm short of it when the elevation is $\alpha$ and goes b cm too far when the elevation is $\beta$. Show that if the velocity of projection is same in all the case, the proper elevation is $\frac{1}{2}{\sin }^{-1}\left[\frac{b\sin 2\alpha +a\sin 2\beta }{a+b}\right]$.

Answer 1

Let the range proper be $D$.

thus the range at angle be $\alpha$ be $D-a$.

Thus,

$D-a=\frac{v^{2}sin2\alpha }{g}...\left(1\right)$

and

The range at angle be $\beta$ be $D+b$.

Thus,

$D+b=\frac{v^{2}sin2\beta }{g}...\left(2\right)$

Let the proper angle of elevation be $\gamma$ .

Thus,

$D=\frac{v^{2}sin2\gamma }{g}...\left(3\right)$

Hence

Multiply (1) by $b$ and (2) by $a$ and add.

$(b+a)D=\frac{v^{2}bsin2\alpha }{g}+\frac{v^{2}asin2\beta }{g}$

Substitute the value of $D$ from (3).

hence,

$(b+a)\frac{v^{2}sin2\gamma }{g}=\frac{v^{2}bsin2\alpha }{g}+\frac{v^{2}asin2\beta }{g}$

Thus,

$sin2\gamma =\frac{bsin2\alpha +asin2\beta }{(b+a)}$

$\gamma =\frac{1}{2}{sin}^{-1}\frac{bsin2\alpha +asin2\beta }{(b+a)}$