Question

A body is projected horizontally from a height of $78.4m$ with a velocity $10m/s$. Its velocity after ${\text{3 seconds}}$ is ($g = 10m/{s^2}$).
 (Take direction of projection as $\vec i$ and vertically upward direction as $\vec j$).
(A) $10\hat i - 30\hat j$
(B) $10\hat i + 30\hat j$
(C) $20\hat i - 30\hat j$
(D) $10\hat i + 10\sqrt {30} \hat j$

Answer 1

Hint: The initial velocity in the vertical direction is zero. The horizontal velocity, however, remains constant throughout.
Formula Used: The formulae used in the solution are given here.
Vertical velocity after time $t$ is given by, ${v_y} = {u_y} - gt$ where $g$ is the acceleration due to gravity and ${u_y}$ is the initial vertical velocity.

Complete Step by Step Solution: For projectile in horizontal projection, consider a projectile, say a ball, thrown horizontally with an initial velocity $\vec u$ from the top of a tower of height $h$.
As the ball moves, it covers a horizontal distance due to its uniform horizontal velocity $\vec u$, and a vertical downward distance because of constant acceleration due to gravity $g$.
It has been given that a body is projected horizontally from a height of $78.4m$ with a velocity $10m/s$.
Initial velocity ${u_x} = 10m/s$.
Initial vertical velocity ${u_y} = 0m/s$.
Vertical velocity after time $t$ is given by, ${v_y} = {u_y} - gt$ where $g$ is the acceleration due to gravity and ${u_y}$ is the initial vertical velocity.
For velocity after ${\text{3 seconds}}$, $t = 3$,
${v_y} = 0 - 10 \times 3 = - 30m/s$.
In the vertically upward direction as $\vec j$, ${v_y} = - 30\hat jm/s$
In the horizontal direction, velocity is $10m/s$.
The velocity of the body after ${\text{3 seconds}}$ is $10\hat i - 30\hat j$.

Hence, the correct answer is Option A.

Note: The height is given to distract the student and is not required in the solution.
A body can be projected in two ways :
Horizontal projection-When the body is given an initial velocity in the horizontal direction only. Angular projection-When the body is thrown with an initial velocity at an angle to the horizontal direction. For horizontal motion, we have constant motion in horizontal because there is no force acting on our object in horizontal direction. Thus, the $X$ component of velocity is constant and acceleration in $X$ direction is zero. The equation that is used to calculate distance and velocity is given below.
$X = V \cdot t$