SATHEE: Kinematics Question 250

Kinematics Question 250

Question: A parachut it’st after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out? [AIEEE 2005]

Options:

A) 293 m

B) 111 m

C) 91 m

D) 182 m

Show Answer

Answer:

Correct Answer: A

Solution:

After bailing out from point A parachut it’st falls freely under gravity. The velocity acquired by it will v From $v^{2} = u^{2} + 2 a s$

$= 0 + 2 \times 9.8 \times 50$ = 980 [As u = 0, $a = 9.8 m / s^{2}$ , s = 50 m] At point B, parachute opens and it moves with retardation of 2 $m / s^{2}$ and reach at ground (Point C) with velocity of $3 m / s$ For the part ?BC? by applying the equation $v^{2} = u^{2} + 2 a s$

$v = 3 m / s$ , $u = \sqrt{980} m / s$ , $a = - 2 m / s^{2}$ , s = h

therefore ${(3)}^{2} = {(\sqrt{980})}^{2} + 2 \times (- 2) \times h$

therefore $9 = 980 - 4 h$

therefore $h = \frac{980 - 9}{4}$

$= \frac{971}{4} = 242.7 \tilde{=} 243$ m. So, the total height by which parachut it’st bail out = $50 + 243$ = 293 m.

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Kinematics Question 250

Question: A parachut it’st after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out? [AIEEE 2005]

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 250

Question: A parachut it’st after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out? [AIEEE 2005]

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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