Answer to Question #107629 in Optics for Faith

Question

A ball bearingX is dropped vertically downwards from the edge of a table and it takes 0.5 seconds to hit the floor.another one y leaves the edge of the table horizontally with a velocity of 5mpr second find
A.time taken for y to reach the floor

Answer 1

As per the given question,

time taken to reach the ground by X one $(t_1)=0.5 sec$

we know that $h=ut+\dfrac{gt^2}{2}$

u=0

$h=0+\dfrac{9.8\times 0.5 \times 0.5}{2}$

$h=1.225m$

now for the Y,

object is projected in horizontal direction, but there is no vertical velocity, so intital velocity in the vertical direction =0

so $h=ut+\dfrac{gt^2}{2}$

$\Rightarrow 1.225=0+\dfrac{9.8\times t^2}{2}$

$t=\sqrt{0.25}=0.5 sec$