SATHEE: Kinematics Question 663

Kinematics Question 663

Question: Two identical particles are projected horizontally in opposite directions with a speed of $5 m s^{- 1}$ each from the top of a tall tower as shown. Assuming $g = 10 m s^{- 2}$ , the distance between them at the moment when their velocity vectors become mutually perpendicular it’s

Options:

A) 2.5 m

B) 5 m

C) 10 m

D) 20 m

Show Answer

Answer:

Correct Answer: B

Solution:

[b] At a time t when velocity vector become mutually perpendicular $v c o s 45^{\circ} = 5$ horizontal component $v = \frac{5}{c o s 45^{\circ}} = 5 \sqrt{2} m/s$ Vertically,

$v \sin 45^{\circ} = g t$

$\Rightarrow t = \frac{v \sin 45^{\circ}}{g} = \frac{5 \sqrt{2} \times 1 / \sqrt{2}}{10} = \frac{5}{10} = \frac{1}{2}$

So, $O A = O B = v \cos 45^{\circ} \times t = 5 \times \frac{1}{2} = 2.5$

$\Rightarrow A B = 2.5 \times 2 = 5 m$

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Kinematics Question 663

Question: Two identical particles are projected horizontally in opposite directions with a speed of 5 ms−1 each from the top of a tall tower as shown. Assuming g = 10 ms−2 , the distance between them at the moment when their velocity vectors become mutually perpendicular it’s

Options:

Answer:

Solution:

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Question: Two identical particles are projected horizontally in opposite directions with a speed of $5 m s^{- 1}$ each from the top of a tall tower as shown. Assuming $g = 10 m s^{- 2}$ , the distance between them at the moment when their velocity vectors become mutually perpendicular it’s