SATHEE: Kinematics Question 619

Kinematics Question 619

Question: A particle is projected with a velocity v such that it’s range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g it’s acceleration due to gravity)

Options:

A) $\frac{4 v^{2}}{5 g}$

B) $\frac{4 g}{5 v^{2}}$

C) $\frac{v^{2}}{g}$

D) $\frac{4 v^{2}}{\sqrt{5} g}$

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Answer:

Correct Answer: A

Solution:

[a] We know, $R = 4 H \cot θ \Rightarrow \cot θ = \frac{1}{2}$

$From triangle we can say that$

$\sin θ = \frac{2}{\sqrt{5}}, cos θ = \frac{1}{\sqrt{5}}$

$∴$ Range of projectile $R = \frac{2 v^{2} \sin θ \cos θ}{g}$

$= \frac{2 v^{2}}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}} = \frac{4 v^{2}}{5 g}$

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Kinematics Question 619

Question: A particle is projected with a velocity v such that it’s range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g it’s acceleration due to gravity)

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Kinematics Question 619

Question: A particle is projected with a velocity v such that it’s range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g it’s acceleration due to gravity)

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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