Question

A stone is thrown horizontally with velocity $\sqrt{2gh}$ from a top of a tower of height $h$ . Find where it will strike the level ground through the foot of the tower. What will be its striking velocity?

Answer 1

Hint: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 and to find the velocity striking add both the velocity.

Formula used:
$s=ut+{\displaystyle \frac{1}{2}}a{t}^2$
Where, $s$ is the displacement of the body, $u$ is the initial velocity, $a$ is the acceleration due to gravity and $t$ is the time taken.

Complete step by step answer:
According to the question, the horizontal velocity of the body is $v=\sqrt{2gh}$.
And we know that,
$s=ut+{\displaystyle \frac{1}{2}}a{t}^2$
The vertical velocity of the body is 0 since there is no vertical movement.
$$\begin{gathered} -h=0-{\displaystyle \frac{1}{2}}g{t}^2 \\ \Rightarrow h={\displaystyle \frac{g{t}^2}{2}} \end{gathered}$$
In order to determine the duration of the movement,
$$\begin{gathered} h={\displaystyle \frac{g{t}^2}{2}} \\ \Rightarrow t=\sqrt{{\displaystyle \frac{2h}{g}}} \end{gathered}$$
Using the formula of distance,
$d=vt$
Now, to find the distance let us substitute the horizontal velocity.
$x=\sqrt{2gh}\times \sqrt{{\displaystyle \frac{2h}{g}}}=2h$
So, the horizontal distance from the tower is obtained as $2h$. Here We see that the modulus of final vertical velocity and horizontal velocity is equal. As such the resultant velocity as stone hits ground will make an angle of ${45}^{\circ }$ with the y-axis so combining both the velocities we get,
$$\begin{gathered} v=\sqrt{{\left(\sqrt{2gh}\right)}^2+{\left(\sqrt{2gh}\right)}^2} \\ \therefore v=2\sqrt{gh} \end{gathered}$$
Hence, the striking velocity is $2\sqrt{gh}$.

Note:Keep in mind that in the above solution, the body makes some horizontal movement before falling to the earth by travelling vertically after the exerted force has passed over it. In the equation of motion, the vertical condition is used to calculate the distance, whereas the horizontal condition is used to find the distance.