Question

An object is projected with velocity ${\vec v_0} = 15\mathop i\limits^ \wedge + 20\mathop j\limits^ \wedge $. Considering x along the horizontal axis and y along the vertical axis. Find its velocity after 2s.

Answer 1

Hint: We know that in physics, the equation of motion is defined as the equations which describe the behavior of a physical system of motion as a function of time. An object in a projectile motion has uniform velocity along the horizontal direction and moves with a uniform acceleration vertically.
Formula used:
 $v - u = at$

Complete answer:
It is given that an object is projected with velocity ${\vec v_0} = 15\mathop i\limits^ \wedge + 20\mathop j\limits^ \wedge $. Now, separating the velocity in ‘x’ and ‘y’ components,
Here, the velocity of the projected object remains the same in the horizontal direction as no acceleration is acting in the horizontal direction in a projectile motion.
So, the velocity along horizontal direction will be same, i.e. $15\mathop i\limits^ \wedge $
Now, we know that the velocity along vertical direction will change.
Thus, $u = 20m/s$, time (t) = 2 seconds, acceleration due to gravity $ = - g = - 9.8m/{s^2}$
The first equation of motion involves the relation between initial and final velocity of an object or a particle, constant acceleration, and time.
Applying the first equation of motion,
$v - u = at$
$ \Rightarrow v - 20 = \left( { - 9.8} \right)2$
$ \Rightarrow v = 20 - 19.6 = 0.4m/s$
This is the velocity of the object in the vertical direction of the projectile motion.
Now, we will write the velocity of the object along horizontal direction and the vertical direction of the projectile motion in vector form,
${v_y} = 0.4\mathop j\limits^ \wedge $ and ${v_x} = 15\mathop i\limits^ \wedge $
So, the velocity of the object at 2 seconds is given as, $15\mathop i\limits^ \wedge + 0.4\mathop j\limits^ \wedge $
Now, the magnitude of the velocity of the object after 2 seconds will be,
$ \Rightarrow \sqrt {{{\left( {15} \right)}^2} + {{\left( {0.4} \right)}^2}} $
$ = \sqrt {\left( {225 + 0.16} \right)} $
$ = \sqrt {225.16} = 15.01m/s$
Hence, the velocity of the projected object after 2 seconds is 15.01 m/s.

Note:
Projectile motion is defined as the motion of an object which is thrown or projected into the air, subjected to only the acceleration due to gravity. Acceleration due to gravity is known as the acceleration gained by an object because of the earth’s gravitational force.