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A projectile is launched at an angle alp...

A projectile is launched at an angle `alpha` from a cliff of height H above the sea level. If it falls into the sea at a distance D from the base of the cliff, show that its maximum height above the sea level is
`[H+(D^(2) tan^(2) alpha)/(4(H+D tan alpha))]`

Text Solution

Verified by Experts

`u cos alphat=D` …(i)
`u cos alpha t-1/2 g t^(2)=-H` …(ii)
`rArr t=(2u sin alpha+-sqrt(u^(2) sin^(2)alpha+2gH))/g=D/(u cos alpha)`
`h=((u sin alpha)^(2))/(2g)=(D^(2)tan^(2)alpha)/(4(H+D tan alpha))`
`:. H_("max")=h+H=H+(D^(2)tan^(2)alpha)/(4(H+D tan alpha))`
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