SATHEE: Kinematics Question 637

Kinematics Question 637

Question: The position of a projectile launched from the origin at $t = 0$ it’s given by $\vec{r} = (40 \hat{i} + 50 \hat{j}) m$ at 2s. If the projectile was launched at an angle $θ$ from the horizontal, then $θ$ it’s (take $g = 10 m s^{- 2}$ )

Options:

A) $\tan^{- 1} \frac{2}{3}$

B) $\tan^{- 1} \frac{3}{2}$

C) $\tan^{- 1} \frac{7}{4}$

D) $\tan^{- 1} \frac{4}{5}$

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Answer:

Correct Answer: C

Solution:

[c] From question,

Horizontal velocity (initial), $u_{x} = \frac{40}{2} =20 m/s$

Vertical velocity (initial), $50= u_{y} t+ \frac{1}{2} g t^{2}$

$\Rightarrow u_{y} \times 2 + \frac{1}{2} (- 10) \times 4; or, u_{y} = \frac{70}{2} = 35 m/s;$

$∴ \tan θ = \frac{u_{y}}{u_{x}} = \frac{35}{20} = \frac{7}{4} \Rightarrow Angle θ = ta n^{- 1} \frac{7}{4}$

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Kinematics Question 637

Question: The position of a projectile launched from the origin at $t = 0$ it’s given by $\vec{r} = (40 \hat{i} + 50 \hat{j}) m$ at 2s. If the projectile was launched at an angle $θ$ from the horizontal, then $θ$ it’s (take $g = 10 m s^{- 2}$ )

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 637

Question: The position of a projectile launched from the origin at t=0 it’s given by r→=(40i^+50j^)m at 2s. If the projectile was launched at an angle θ from the horizontal, then θ it’s (take 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: The position of a projectile launched from the origin at $t = 0$ it’s given by $\vec{r} = (40 \hat{i} + 50 \hat{j}) m$ at 2s. If the projectile was launched at an angle $θ$ from the horizontal, then $θ$ it’s (take $g = 10 m s^{- 2}$ )