Question

A ball is projected horizontally. After $3\;\sec$ from projection its velocity becomes $1.25$ times the velocity of the projection. Its velocity of projection is:
A) $10\;{\rm{m/s}}$
B) $20\;{\rm{m/s}}$
C) $30\;{\rm{m/s}}$
D) $40\;{\rm{m/s}}$

Answer 1

The ball is projected in horizontal direction hence after $3\;\sec$ the horizontal component of the velocity remains constant while the y component becomes zero initially but it will change linearly.

Complete step by step answer:
Let us assume that the horizontal component of the velocity is ${v_x}$.
The projectile motion is the form of two-dimensional motion in which an object is thrown under the action of gravity. The projectile follows a parabolic path.
We can calculate the y component of velocity with the help of the first equation of motion.
${v_y} = u + at$
Here, the initial velocity is $u$, the acceleration is $a$ and the time is $t$ and the y component of velocity is ${v_y}$.
We will now substitute the known values in the above equation of motion.
$\Rightarrow {v_y} = 0 + 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \times 3\,{\rm{s}}$
On simplification,
$\Rightarrow {v_y} = 0 + 29.43\;{\rm{m/s}}$
$\Rightarrow {v_y} = {\rm{29}}{\rm{.43}}\;{\rm{m/s}}$
The relation between the velocity and x component of velocity is given by $v = 1.25{v_x}$.
Now we have both x and y components of velocity therefore, the velocity can be calculated by the use of Pythagoras theorem.
$v = \sqrt {v_x^2 + v_y^2}$
We will now substitute the known values in the above equation of velocity.
$\Rightarrow 1.25{v_x} = \sqrt {v_x^2 + {{\left( {29.43} \right)}^2}}$
In the question, it is given that the velocity becomes 1.25 times the horizontal component of the velocity after three seconds.
$\Rightarrow 1.25{v_x} = \sqrt {v_x^2 + {{\left( {29.43} \right)}^2}}$
$\Rightarrow 1.5625v_x^2 = v_x^2 + {\left( {29.43} \right)^2}$
On simplification,
$\Rightarrow 0.5625v_x^2 = {\rm{866}}{\rm{.1249}}$
$\Rightarrow v_x^2 = \dfrac{{866.1249}}{{0.5625}}$
We can further solve the above equation,
$\Rightarrow {v_x} = \sqrt {\dfrac{{866.1249}}{{0.5625}}}$
$\Rightarrow {v_x} = \sqrt {1539.8}$
On further simplification,
$\Rightarrow {v_x} = {\rm{39}}{\rm{.24}}\;{\rm{m/s}}$
$\Rightarrow {v_x} \approx 40\;{\rm{m/s}}$

Thus, the velocity of projection is calculated to be $40\;{\rm{m/s}}$ and thus from the given options, only option D is correct.

Note:
Make sure not to get confused between the velocity of projection and x component of the velocity; these both are the same term. The correct use of the Pythagoras theorem will lead you to get the final answer.