A projectile crosses two walls of equal height H symmetrically as shown angle of projection of proje

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3-2.Motion in Plane

hard

A projectile crosses two walls of equal height $H$ symmetrically as shown The angle of projection of the projectile is

$tan^{-1}(3/4)$

$tan^{-1}(4/3)$

$tan^{-1}(4/5)$

$tan^{-1}(3/5)$

The projectile is at a height $h$ at $t =2$ and $t =6$ seconds .

$h=u \sin \theta t-\frac{1}{2} g t^2$

at $t =2$,

$h=2 u \sin \theta-2 g$

at $t=6$,

$h =6 u \sin \theta-18\,g$

$\therefore 4 u \sin \theta=16\,g$

$u \sin \theta=4 g =40 \ldots \text { (i) }$

As the separation between the 2 points is $120 m$,

$u \cos \theta \times \Delta t =120$

$4 u \cos \theta=120$

$u \cos \theta=30 \ldots$ $(ii)$

Dividing equation $(i)$ by $(iii)$

$\tan \theta=\frac{4}{3}$

Standard 11

Physics

medium

medium

hard

hard

medium