A body is thrown horizontally with velocity √(2gh) from the top of a tower of height h. It strikes the level ground through the foot of tower at a distance x from the tower. The value of x is

The correct option is 2h.

To find the horizontal distance $x$ where the body strikes the ground, we can analyze the horizontal and vertical motions of the body independently.

1. Vertical Motion:
The body is thrown horizontally, which means its initial vertical velocity ($u_y$) is zero.
The vertical distance traveled is the height of the tower, $h$.
Using the second equation of motion for the vertical direction:
$h=u_{y}t+\frac{1}{2}g{t}^2$
Since $u_y=0$, this simplifies to:
$h=\frac{1}{2}g{t}^2$
Solving for the time of flight ($t$):
$t^2=\frac{2h}{g}$
$t=\sqrt{\frac{2h}{g}}$

2. Horizontal Motion:
There is no horizontal acceleration acting on the body (ignoring air resistance). Therefore, the horizontal velocity remains constant throughout the flight.
Given horizontal velocity, $u=\sqrt{2gh}$
The horizontal distance ($x$) traveled by the body in time $t$ is:
$x=u\times t$
Substituting the values of $u$ and $t$:
$x=\sqrt{2gh}\times \sqrt{\frac{2h}{g}}$
$x=\sqrt{2gh\times \frac{2h}{g}}$
The acceleration due to gravity ($g$) cancels out:
$x=\sqrt{4h^2}$
$x=2h$

Thus, the horizontal distance from the tower where the body strikes the ground is $2h$.