SATHEE: Kinematics Question 512

Kinematics Question 512

Question: A man standing on the roof of a house of height $h$ throws one particle vertically downwards and another particle horizontally with the same velocity u. The ratio of their velocities when they reach the earth’s surface will be

Options:

A) $\sqrt{2 g h + {u_{^{^{}}}}^{2}} : u$

B) $1 : 2$

C) $1 : 1$

D) $\sqrt{2 g h + u^{2}} : \sqrt{2 g h}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] When particle is thrown in vertical downward direction with velocity u, then the final velocity at the ground level it’s $v^{2} = u^{2} + 2 g h$

$∴ v = \sqrt{u^{2} + 2 g h}$

Another particle is thrown horizontally with same velocity then at the surface of earth.

Horizontal component of velocity $v_{x} = u$

Therefore, resultant velocity, $v = \sqrt{u^{2} + 2 g h}$

For both the particles, final velocities when they reach the earth’ surface are equal.

Kinematics Question 512

Question: A man standing on the roof of a house of height $h$ throws one particle vertically downwards and another particle horizontally with the same velocity u. The ratio of their velocities when they reach the earth’s surface will be

Options:

Answer:

Solution:

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Mentorship

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Kinematics Question 512

Question: A man standing on the roof of a house of height h throws one particle vertically downwards and another particle horizontally with the same velocity u. The ratio of their velocities when they reach the earth’s surface will be

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: A man standing on the roof of a house of height $h$ throws one particle vertically downwards and another particle horizontally with the same velocity u. The ratio of their velocities when they reach the earth’s surface will be