SATHEE: Kinematics Question 176

Kinematics Question 176

Question: A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact it’s [BHU 1997; CPMT 1997]

Options:

A) 2100 $m / \sec^{2}$ downwards

B) 2100 $m / \sec^{2}$ upwards

C) 1400 $m / \sec^{2}$

D) 700 $m / \sec^{2}$

Show Answer

Answer:

Correct Answer: B

Solution:

Velocity at the time of striking the floor,

$u = \sqrt{2 g h_{1}} = \sqrt{2 \times 9.8 \times 10} = 14 m / s$ Velocity with which it rebounds.

$v = \sqrt{2 g h_{2}} = \sqrt{2 \times 9.8 \times 2.5} = 7 m / s$

$∴$ Change in velocity $Δ v = 7 - (- 14) = 21 m / s$

$∴$ Acceleration $= \frac{Δ v}{Δ t} = \frac{21}{0.01} = 2100 m / s^{2}$ (upwards)

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Kinematics Question 176

Question: A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact it’s [BHU 1997; CPMT 1997]

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 176

Question: A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact it’s [BHU 1997; CPMT 1997]

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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