Q. A glass marble projected horizontally from the top of a table falls at a distance $x$ from the edge of the table. If $h$ is the height of the table, then the velocity of projection is
A) $h\sqrt{\frac{g}{2x}}$
B) $x\sqrt{\frac{g}{2h}}$
C) $gxh$
D) $gx + h$
Explanation
(b) $t = \sqrt{\frac{2h}{g}},\ x = v\sqrt{\frac{2h}{g}}\ \ or\ \ v = x\sqrt{\frac{g}{2h}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.