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Show that maximum horizontal range is 4 ...

Show that maximum horizontal range is 4 times the maximum height attained by the projectile.

Text Solution

Verified by Experts

We know that
`R = ( u ^(2) sin 2 theta)/( g ) and H = ( u ^(2) sin ^(2) theta)/(2g)`
Range is maximum , when `theta = 45 ^(@),` therefore
`R _(max) = ( u ^(2) sin ( 2 xx 45 ^(@)))/( g ) and H = ( u ^(2) sin ^(2) 45 ^(@))/( g ) or R _(max) = ( u ^(2))/(g) and H = ( u ^(2))/( 4g)`
Dividing both, we get
`(R _(max))/(H) =( u ^(2))/( g ) xx ( 4g)/( u ^(2)) = 4 or R _(max) = 4 H.` Hence proved.
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