A tank open at the top with a water level of 1 m, as shown

A tank open at the top with a water level of 1 m, as shown in the figure, has a hole at a height of 0.5 m. A free jet leaves horizontally from the smooth hole. The distance X(in m) where the jet strikes the floor is

u_x = √2gh

= √2 × 9.8 × 0.5
u_x = 3.13 m/s
Now for distance x
we require u_x so–

x = u_x . t +	1	× a × b
	2

x = 3.13 t
(∵ acceleration (a) = 0 in horizontal direction)

Now for t ⇒ y = u_yt +	1	gt²
	2

⇒ 0.5 = 0 +	1	× 9.81 × t² [∵ u_y = 0]
	2

t = 0.319 sec.
so distance, x = 3.13 × 319 = 1m