Question

A body is thrown horizontally with a velocity $\sqrt {2gh}$ from the top of a tower of height $h$ . It strikes the level ground through the foot of the tower at a distance $x$ from the tower. The value of $x$ is:
A) $h$
B) $\dfrac{h}{2}$
C) $2h$
D) $\dfrac{{2h}}{3}$

Answer 1

Use the displacement formula in the equation of motion, substitute the vertical travelling condition in it to find the time taken. Substitute the obtained time taken in the formula of the distance to find the distance of the falling body from the tower.

Formula used:
(1) The equation of the motion is given by
$s = ut + \dfrac{1}{2}a{t^2}$
Where $s$ is the displacement of the body, $u$ is the initial velocity of the body, $a$ is the acceleration of the body and $t$ is the time taken for the displacement.
(2) The formula of the displacement is given by
$x = vt$
Where $x$ is the displacement of the body, $v$ is the velocity of the body and $t$ is the time taken for the movement.

Complete step by step solution:
The horizontal velocity of the body, $v = \sqrt {2gh}$
Let us use the formula of the equation of the motion.
$s = ut + \dfrac{1}{2}a{t^2}$
Since there is no vertical movement, the vertical velocity of the body is zero.
$\Rightarrow - h = 0 - \dfrac{1}{2}g{t^2}$
By simplification of the above step, we get
$\Rightarrow h = \dfrac{{g{t^2}}}{2}$
In order to find the time taken for the movement,
$\Rightarrow t = \sqrt {\dfrac{{2h}}{g}}$
Using the formula of the distance,
$d = vt$
Substitute the horizontal velocity to find the distance from the tower that the body falls from the tower.
$\Rightarrow x = \sqrt {2gh} \times \sqrt {\dfrac{{2h}}{g}} = 2h$
The horizontal distance that the body falls away from the tower is obtained as $2h$ .

Thus the option (C) is correct.

Note: In the above solution, keep in mind that the body makes some horizontal movement and then after the exerted force gats over it falls on the ground by travelling vertically. So the vertical condition is applied in the equation of motion and the horizontal condition is applied to find the distance.