Question

A body is projected with a velocity of$4\sqrt 2 m{s^{ - 1}}$. The velocity of the body after $0.7s$ will be nearly: (Take$g = 10m{s^{ - 2}}$)
(A) $10m{s^{ - 1}}$
(B) $9m{s^{ - 1}}$
(C) $19m{s^{ - 1}}$
(D) $11m{s^{ - 1}}$

Answer 1

Hint We are given with the velocity of the projected body and are asked about the velocity of the body after a given amount of time. Thus, we will apply the appropriate equation of motion. Finally, we will choose a correct option out of the four given options.
Formulae Used:
$v = u + at$
Where,$v$ is the final velocity of the body,$u$ is the initial velocity of the body,$a$ is the acceleration of the body and$t$ is the time of travel of the body.
${v_{net}} = \sqrt {{{({v_x})}^2} + {{({v_y})}^2}} $
Where,${v_{net}}$ is the net velocity of a body,${v_x}$ is the velocity of the body in the x direction and${v_y}$ is the velocity of the body in the y direction.

Complete Step By Step Solution
Here,
The initial projected velocity,$u = 4\sqrt 2 m{s^{ - 1}}$
Acceleration due to gravity,$g = 10m{s^{ - 2}}$
Time of travel,$t = 0.7s$
Now,
When the projectile is projected and falls under the effect of gravity.
Then, the projectile will approach downwards.
Thus, the direction of the acceleration due to gravity and the approach of the projectile is the same.
Thus, the sign of the acceleration due to gravity is positive.
Now,
Transforming the motion equation into a gravitational form.
Thus,
$v = u + gt$
Now,
For a projectile, we will separate the velocity into x and y directions.
Now,
The velocity in the x direction, ${v_x} = u = 4\sqrt 2 m{s^{ - 1}}$
The velocity in the y direction,${v_y} = gt$
Thus,
Substituting this values, we get
${v_y} = (10)(0.7) = 7m{s^{ - 1}}$
Now,
For the net velocity of the projectile, we get
${v_{net}} = \sqrt {{{(4\sqrt 2 )}^2} + {{(7)}^2}} $
Now,
Calculating the values, we get
${v_{net}} = \sqrt {32 + 49} $
Now,
Further, we get
${v_{net}} = \sqrt {81} $
Finally, we get
${v_{net}} = 9m{s^{ - 1}}$

Hence, we get the correct option as (B).

Note We have separated the velocity into the x and y direction as the trajectory of the motion is dependent on the two directions. If the trajectory only depended on only one direction, then we do not need to separate into the two directions and should only consider the depended trajectory.