SATHEE: Kinematics Question 530

Kinematics Question 530

Question: A particle is projected from the ground with an initial speed of $v$ at an angle $θ$ with horizontal. The average velocity of the particle between it’s point of projection and highest point of trajectory it’s

Options:

A) $\frac{v}{2} \sqrt{1 + 2 \cos^{2} θ}$

B) $\frac{v}{2} \sqrt{1 + 2 \cos^{2} θ}$

C) $\frac{v}{2} \sqrt{1 + 3 \cos^{2} θ}$

D) $v \cos θ$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Average velocity $= \frac{D i s p l a c e m e n t}{T i m e}$

$= \frac{\sqrt{H^{2} + R^{2} / 4}}{T / 2}$

Putting the required values, we get $v_{a v} = \frac{v}{2} \sqrt{1 + 3 \cos^{2} θ}$

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Kinematics Question 530

Question: A particle is projected from the ground with an initial speed of v at an angle θ with horizontal. The average velocity of the particle between it’s point of projection and highest point of trajectory it’s

Options:

Answer:

Solution:

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Question: A particle is projected from the ground with an initial speed of $v$ at an angle $θ$ with horizontal. The average velocity of the particle between it’s point of projection and highest point of trajectory it’s