A body is thrown with the velocity $u$ at an angle $θ$ with the horizontal. If the body remains in air for 6 seconds, the maximum height reached by the body will be

9.8 m

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19.6 m

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20.0 m

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44.1 m

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Solution

The correct option is D $44.1 m$
$T = \frac{2 u sin θ}{g} = 6$
$u sin θ = 3 g$
$H = \frac{u^{2} {sin}^{2} θ}{2 g} = \frac{9 g^{2}}{2 g} = 4.5 g$
$= 44.1 m$

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A body is thrown with the velocity u at an angle θ with the horizontal. If the body remains in air for 6 seconds, the maximum height reached by the body will be

The correct option is D 44.1 mT=2u sin θg=6 u sin θ=3g H=u2sin2 θ2g=9g22g=4.5 g =44.1 m

A body is thrown with the velocity $u$ at an angle $θ$ with the horizontal. If the body remains in air for 6 seconds, the maximum height reached by the body will be

The correct option is D $44.1 m$
$T = \frac{2 u sin θ}{g} = 6$
$u sin θ = 3 g$
$H = \frac{u^{2} {sin}^{2} θ}{2 g} = \frac{9 g^{2}}{2 g} = 4.5 g$
$= 44.1 m$