For an object thrown at $$45$$° to the horizontal, the maximum height H and horizontal range R are related as

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Infinity Learn · Accepted Answer

As we know that, maximum height of a projectile is given $$byHmax=u2sin2$$θ$$2gwhere$$, u $$=$$ initial velocity of projectile            g $$=$$ acceleration due to gravityand      θ $$=$$ angle of projection.As per question,$$Hmax=v2sin24502g$$                  ... (i) (As$$, u $$=$$ v, θ $$=$$ $$45$$°$$)Now$$, range of a projectile is given byR $$=u2$$ $$sin$$ $$2$$θg⇒  R $$=v2$$ $$sin$$ $$(2$$ × $$450)g$$⇒  R $$=v2$$ $$sin$$ $$900g$$              ... $$(ii)On$$ dividing Eq. $$(i) by Eq. (ii), we $$getHmaxR=v2sin245$$°×$$g2g$$×$$v2sin90$$°$$=14$$×$$1$$⇒  $$R=4Hmax=4H$$

For an object thrown at 45° to the horizontal, the maximum height H and horizontal range R are related as

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