A projectile crosses two walls of equal heigh | vedclass

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Question

The projectile is at a height $h$ at $t =2$ and $t =6$ seconds .

$h=u \sin 	heta t-\frac{1}{2} g t^2$

at $t =2$,

$h=2 u \sin 	heta-2 g$

Vedclass · Accepted Answer

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

Vedclass · Answer

The projectile is at a height $h$ at $t =2$ and $t =6$ seconds .

$h=u \sin 	heta t-\frac{1}{2} g t^2$

at $t =2$,

$h=2 u \sin 	heta-2 g$

at $t=6$,

$h =6 u \sin 	heta-18\,g$

$	herefore 4 u \sin 	heta=16\,g$

$u \sin 	heta=4 g =40 \ldots 	ext { (i) }$

As the separation between the 2 points is $120 m$,

$u \cos 	heta 	imes \Delta t =120$

$4 u \cos 	heta=120$

$u \cos 	heta=30 \ldots$ $(ii)$

Dividing equation $(i)$ by $(iii)$

$	an 	heta=\frac{4}{3}$

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

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