SATHEE: Kinematics Question 186

Kinematics Question 186

Question: A body is released from the top of a tower of height $h$ . It takes $v = \frac{1}{2} b t^{2} + v_{0}$ sec to reach the ground. Where will be the ball after time $t / 2$ sec [NCERT 1981; MP PMT 2004]

Options:

A) At $h / 2$ from the ground

B) At $h / 4$ from the ground

C) Depends upon mass and volume of the body

D) At $3 h / 4$ from the ground

Show Answer

Answer:

Correct Answer: D

Solution:

Let the body after time $t / 2$ be at x from the top, then $x = \frac{1}{2} g \frac{t^{2}}{4} = \frac{g t^{2}}{8}$ -(i) $h = \frac{1}{2} g t^{2}$ -(ii) Eliminate t from (i) and (ii),

we get $x = \frac{h}{4}$

$∴$ Height of the body from the ground $= h - \frac{h}{4} = \frac{3 h}{4}$

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अगला

Kinematics Question 186

Question: A body is released from the top of a tower of height $h$ . It takes $v = \frac{1}{2} b t^{2} + v_{0}$ sec to reach the ground. Where will be the ball after time $t / 2$ sec [NCERT 1981; MP PMT 2004]

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 186

Question: A body is released from the top of a tower of height h . It takes v=12bt2+v0 sec to reach the ground. Where will be the ball after time t/2 sec [NCERT 1981; MP PMT 2004]

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: A body is released from the top of a tower of height $h$ . It takes $v = \frac{1}{2} b t^{2} + v_{0}$ sec to reach the ground. Where will be the ball after time $t / 2$ sec [NCERT 1981; MP PMT 2004]