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physics

mechanics

Easy

Question

In previous problem, the angular momentum of the projectile about the origin when it just before hits the ground should be

$\frac{- 2 m v_{i}^{3} \sin θ \sin θ}{g} \hat{k}$
$\frac{m v_{i}^{3} \sin θ^{2} \cos θ}{2 g} \hat{k}$
$\frac{- m v_{i}^{3} \sin θ^{2} \cos θ}{g} \hat{k}$
$\frac{2 m v_{i}^{3} \sin θ^{2} \cos θ}{g} \hat{k}$

The correct answer is: $\frac{- 2 m v_{i}^{3} \sin θ \sin θ}{g} \hat{k}$

$L with not stretchy rightwards arrow on top subscript 2 equals R i with not stretchy hat on top cross times m v with not stretchy rightwards arrow on top subscript 2 comma text where end text R equals fraction numerator v subscript i superscript 2 sin invisible function application 2 theta over denominator g end fraction equals fraction numerator v subscript i superscript 2 left parenthesis 2 sin invisible function application theta cos invisible function application theta right parenthesis over denominator g end fraction equals negative m R v subscript i sin invisible function application theta k with not stretchy hat on top equals fraction numerator negative 2 m v subscript i superscript 3 sin invisible function application theta sin invisible function application theta over denominator g end fraction k with not stretchy hat on top$

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$text A projectile of mass end text straight m text is launched with an initial velocity end text v with not stretchy rightwards arrow on top subscript i text making an angle end text theta text with the end text text horizontal as shown in figure. The projectile moves in the gravitational field of the Earth. The end text text angular momentum of the projectile about the origin when it is at the highest point of its end text text trajectory is end text$

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PhysicsMechanics

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