SATHEE: Kinematics Question 636

Kinematics Question 636

Question: An object it’s projected with a velocity of $20 m / s$ making an angle of $45^{\circ}$ with horizontal. The equation for the trajectory it’s $h = A x - B x^{2}$ where h it’s height, x it’s horizontal distance, A and B are constants. The ratio A: b is $(g = 10 m s^{- 2})$

Options:

A) $1 : 5$

B) $5 : 1$

C) $1 : 40$

D) $40 : 1$

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Given $h = A x - B x^{2}$ ,

on comparing with $y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$ ,

we get $A = \tan θ = t a n 45^{\circ} = 1,$

and $B = \frac{g}{2 u^{2} \cos^{2} θ} = \frac{10}{2 \times 20^{2} \times \cos^{2} 45^{\circ}} = \frac{1}{40}$

$∴ \frac{A}{B} = 40.$

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Kinematics Question 636

Question: An object it’s projected with a velocity of $20 m / s$ making an angle of $45^{\circ}$ with horizontal. The equation for the trajectory it’s $h = A x - B x^{2}$ where h it’s height, x it’s horizontal distance, A and B are constants. The ratio A: b is $(g = 10 m s^{- 2})$

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 636

Question: An object it’s projected with a velocity of 20m/s making an angle of 45∘ with horizontal. The equation for the trajectory it’s h=Ax−Bx2 where h it’s height, x it’s horizontal distance, A and B are constants. The ratio A: b is (g = 10 ms−2)

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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Question: An object it’s projected with a velocity of $20 m / s$ making an angle of $45^{\circ}$ with horizontal. The equation for the trajectory it’s $h = A x - B x^{2}$ where h it’s height, x it’s horizontal distance, A and B are constants. The ratio A: b is $(g = 10 m s^{- 2})$