Question

For a projection, $(range{)}^2$ is $48$ times of $($maximum height${)}^2$ obtained. Find angle projection.

Answer 1

Let angle of projection $=a^{\circ }$ with horizontal.

Initial velocity$y=u$

vertical component of velocity$y=u.\sin a$.

Horizontal component of velocity$y=u\cos a$.

Max.height =h

Range =R

Vertical motion:

$v^2=u^2-2gh$

$(0{)}^2=(u.sina{)}^2-2gh$

$h=u^2.si{n}^{2}a/2g\dots \dots \dots \dots \dots ..\left(2\right)$

$v=u-g.t$

$0=usina-gt$

$t=u.sina/g$

Total time of flight

$\left(T\right)=2.t.=2u.\frac{\sin a}{g}\dots \dots ..\left(3\right)$

Horizontal motion:-

Range(R)=horizontal component of velocity$\times$T

$R=ucosa\times 2u.\frac{\sin a}{g}$

$R=u^2.\frac{(2\sin a.\cos a)}{g}\dots \dots \dots \dots \dots \dots .\left(4\right)$

$R^2=48.h^2$

$u^4.\frac{(4.si{n}^{2}a.co{s}^{2}a)}{g^2}=48.u^4.\frac{si{n}^{4}a}{4g^2}$

$co{s}^{2}a=3.si{n}^{2}a.$

$\frac{1}{3}=\frac{si{n}^{2}a}{co{s}^{2}a}$

${\tan }^{2}a=\frac{1}{3}$

$\tan a=\frac{1}{3^1/2}=\tan 30°$

or $a=30°$