A projectile is thrown in upward direction making an angle of 60^o with horizontal direction with a

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3-2.Motion in Plane

hard

A projectile is thrown in the upward direction making an angle of $60^o$ with the horizontal direction with a velocity of $150\, ms^{-1}$. Then the time after which its inclination with the horizontal is $45^o$ is

$15\,\left( {\sqrt 3 - 1} \right)\,s$

$15\,\left( {\sqrt 3 + 1} \right)\,s$

$7.5\,\left( {\sqrt 3 - 1} \right)\,s$

$7.5\,\left( {\sqrt 3 + 1} \right)\,s$

Solution

$\mathrm{v} \cos 45^{\circ}=150 \cos 60^{\circ} \quad \Rightarrow \mathrm{v}=75 \sqrt{2} \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}+\mathrm{a}_{\mathrm{y}} \mathrm{t} \quad \Rightarrow \mathrm{v} \sin 45^{\circ}=150 \sin 60^{\circ}-\mathrm{g} \times \mathrm{t}$

$\mathrm{t}=\frac{150 \sin 60^{\circ}-\mathrm{v} \sin 45^{\circ}}{10}=7.5(\sqrt{3}-1) \sec$

Standard 11

Physics

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(AIIMS-2019)

A projectile is thrown in the upward direction making an angle of $60^o$ with the horizontal direction with a velocity of $150\, ms^{-1}$. Then the time after which its inclination with the horizontal is $45^o$ is

Solution

Similar Questions

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