Question

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Answer 1

Answer:

(4) 3R is the answer

Step-by-step explanation:

Given :

$\mathrm{u = \sqrt{2gR}}$

To find :

The horizontal distance between centre 'O' and the point where it strikes the ground

Solution :

Let us find the time required for the body to reach the ground

We know that,

$\boxed{\mathrm{t = \sqrt{\dfrac{2h}{g}}}}$

Here, h = height = R

Thus,

$\implies \mathrm{t = \sqrt{\dfrac{2R}{g}}}$

Now, Let us find the horizontal distance when it strikes the ground at time t

We know that,

$\boxed{\mathrm{s_x = u_xt + \dfrac{1}{2}gt^2}}$

$\implies \mathrm{s_x = \sqrt{2\not g \ R}\sqrt{\dfrac{2R}{\not g}} + \dfrac{1}{2}g\bigg(\sqrt{\dfrac{2R}{g}}\bigg)^2}$

$\implies \mathrm{s_x = 2R + \dfrac{1}{\not2}\not g\dfrac{\not2 R}{\not g}}$

$\implies \mathrm{s_x = 2R + R}$

$\therefore \underline{\boxed{\mathbf{s_x = 3R}}}$

Hope it helps!!