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Standard XII

Physics

Direction of Instantaneous Velocity

Particle is d...

Question

Particle is dropped from the height of 20m on horizontal ground. There is wind blowing due to which horizontal acceleration of the particle becomes 6m $s^{- 2}$ . Find the horizontal displacement of the particle till it reaches ground.

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10m

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12m

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24m

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Solution

The correct option is C 12m
The correct opption is C.

Given,

$h e i g h t = 20 m$

$A c c e l e r a t i o n = 6 m / s^{2}$

We know $u = 0, g = 10 m / s^{2}, s = 20 m$

So,

$s = u t + \frac{1}{2} a t^{2}$

$s = \frac{1}{2} g t^{2}$

$20 = \frac{1}{2} 10 t^{2}$

$t = 2 s$

Since $a = 6 m / s^{2}$ in the horizonttal,

Thus,

$s = u t + \frac{1}{2} a t^{2}$

$s = \frac{1}{2} a t^{2}$

$= \frac{1}{2} g \times 2^{2}$

$= 12 m$

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Particle is dropped from the height of 20m on horizontal ground. There is wind blowing due to which horizontal acceleration of the particle becomes 6ms−2. Find the horizontal displacement of the particle till it reaches ground.

The correct option is C 12mThe correct opption is C.Given,height=20mAcceleration=6m/s2We know u=0,g=10m/s2,s=20mSo,s=ut+12at2s=12gt220=1210t2t=2sSince a=6m/s2 in the horizonttal,Thus,s=ut+12at2s=12at2=12g×22=12m

Particle is dropped from the height of 20m on horizontal ground. There is wind blowing due to which horizontal acceleration of the particle becomes 6m $s^{- 2}$ . Find the horizontal displacement of the particle till it reaches ground.