Refer to the figure. The speed of the particle when its velocity becomes parallel to the inclined plane is

$5 m / s$
$5 (\sqrt 3 + 1) / 2 \sqrt 3 m / s$
$5 \sqrt 3 m / s$
$5 (\sqrt 3 - 1) m / s$

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Solution

The correct option is D
$5 (\sqrt 3 - 1) m / s$

Step 1: Given data
From the above figure
$u =$ $10 m / s$
We have to find the speed of the particle.
Step 2: Concept to be used
Velocity is defined as the distance travelled by the body in unit time in a given direction.
Step 3: Find the speed of the particle
As taking $V_{x}$ as the velocity of the particle when velocity becomes parallel to the inclined plane,
$V_{x} = u_{x} + a_{x} \times t V_{x} = u \cos 30 ° - g \sin 45 ° \times t$
$= 10 \times \frac{\sqrt{3}}{2} - 10 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$= 5 \sqrt{3} - 5$
Factor out common term $5$ .
Factor of $5 = 5 \times 1$
So,
$V_{x} = 5 (\sqrt{3} - 1)$
Hence, the correct option is D.

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Refer to the figure. The speed of the particle when its velocity becomes parallel to the inclined plane is5m/s5(√3+1)/2√3m/s5√3m/s5(√3–1)m/s

Refer to the figure. The speed of the particle when its velocity becomes parallel to the inclined plane is

$5 m / s$
$5 (\sqrt 3 + 1) / 2 \sqrt 3 m / s$
$5 \sqrt 3 m / s$
$5 (\sqrt 3 - 1) m / s$