Question

A body is thrown horizontally from a tower, 100m high with a velocity $40{\text{ }}m/{s^{ - 1}}$ . Find the time ( in a sec) when it is moving at an angle of $40^\circ$ with horizontal. (Take $g{\text{ }} = {\text{ }}10{\text{ }}m/{s^{ - 2}}$ )

Answer 1

-The body is moving at an angle with the horizontal direction. Its motion is like a projectile motion. There are no horizontal components of acceleration. Hence the velocity along the horizontal direction remains constant.
-the horizontal and the vertical components of velocity are equal due to the angle.
-Find the time from the velocity component along the vertical direction. Here the initial velocity is zero.

Formula used:
The component of velocity along $'X'$ axis is ${v_x}$ and The component of velocity along $'Y'$ axis is ${v_y}$.
$\tan \theta = \dfrac{{{v_x}}}{{{v_y}}}$
${v_x}$ is uniform velocity. (value is given).
${v_y} = gt$ [since the initial velocity is zero]
Where $t$ is the required time and $g$ is the gravitational acceleration

Complete step by step answer:
The body which is thrown from a tower horizontally is moving at an angle of $45^\circ$ with the horizontal direction.
This is like a projectile motion. So, there are no components of acceleration along the $X$ axis. Hence the velocity along the $X$ axis is uniform.
The component of velocity along $'X'$ axis is ${v_x}$ and The component of velocity along $'Y'$ axis is ${v_y}$.
Given, ${v_x} = 40m/s$
$\tan \theta = \dfrac{{{v_x}}}{{{v_y}}}$ where, $\theta = 45^\circ$
$\therefore \tan 45^\circ = \dfrac{{{v_x}}}{{{v_y}}}$
$\Rightarrow {v_x} = {v_y}$ $[\because \tan 45^\circ = 1]$
$\therefore {v_y} = 10m/s$
The equation of motion for the vertical direction i.e. along $'Y'$ axis is, ${v_y} = gt$ [since the initial velocity is zero]
Where $t$ is the required time and $g$ is the gravitational acceleration
Given, $g = 10m/{s^2}$
$\therefore t = \dfrac{{{v_y}}}{g}$
$\Rightarrow t = \dfrac{{40}}{{10}}$
$\Rightarrow t = 4s$

Hence the time when the body is moving, $t = 4s$.

Note:
-The gravitational acceleration is acting downwards i.e vertically. Hence it has no components along the horizontal direction.
Here, $'X'$ denotes the horizontal axis and $'Y'$ denotes the vertical axis .
-So there are no components of acceleration along the $'X'$ axis. Hence the velocity remains the same from the initial position to the final position. That’s why, The component of velocity along $'X'$ axis , ${v_x}$ is a uniform velocity.
-We assume that the body is at rest initially. So we take the initial velocity vertically zero.