Question

Hey! Any physics experts? Kindly solve this one.

A small block of mass m is projected horizontally from the top of the
smooth and fixed hemisphere of radius r with speed u as shown. For values of u >=uo, (uo = Vgr ) it does not slide on the hemisphere. [ i.e. leaves the surface at the top itself ]. For u = 2 uo, it lands at point P on ground. Find OP.
(A) 2
(B) 2r
(C) 4r
(D) 22r​

Answer 1

Answer:

$(D) 2\sqrt{2}$

Explanation:

1)  We have,

$S_y=-r\\ \\ u_y=0,a_y=-g$

Now, Using Newton's equation of  motion,

$S_y=u_yt+\frac{1}{2}a_yt^2\\ \\=>-r=0*t-\frac{1}{2}gt^2\\ \\=>-r=\frac{-gt^2}{2}\\ \\=>t=\sqrt{\frac{2r}{g}}$

That, Time taken by block to reach the ground is $t=\sqrt{\frac{2r}{g}}$

2) Now, Distance covered by block is given by

$OP = u_xt=2u_0t\\ \\ =2 \sqrt{gr} \sqrt{\frac{2r}{g}}\\ \\=2\sqrt{2}r$

Hence, Option (D) is right answer .

Answer 2

Explanation:

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