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**Hint:**The differentiation of position with respect to time is equal to velocity and the both the velocities are perpendicular to each other. Find the magnitude of both the velocities.

**Formula used:**The velocity of a body is given by $v = \dfrac{{dx}}{{dt}}$ where $x$ is the position or displacement of the body and $t$ is time.

**Complete step by step answer:**

From the given question, we know that the vertical height of the projectile is $y = \left( {4t} \right)\;{\rm{m}}$ and the horizontal distance of the projectile is $x = \left( {3t} \right)\;{\rm{m}}$.

We know that the differentiation of position with respect to time is equal to velocity.

The vertical velocity of the projectile is,

${v_y} = \dfrac{{dy}}{{dt}}$

Substitute the expression in the above equation. We get,

$

{v_y} = \dfrac{{d\left( {4t} \right)}}{{dt}}\\

= 4\;{\rm{m/s}}

$

Similarly, the vertical velocity of the projectile is,

${v_x} = \dfrac{{dx}}{{dt}}$

Substitute the expression in the above equation. We get,

$

{v_x} = \dfrac{{d\left( {3t} \right)}}{{dt}}\\

= 3\;{\rm{m/s}}

$

The velocity of the of a projectile on a certain planet is,

$v = \sqrt {v_x^2 + v_y^2} $

Substitute the values in the above equation. We get,

\[

v = \sqrt {{3^2} + {4^2}} \\

= \sqrt {9 + 16} \\

= 5\;{\rm{m/s}}

\]

We have to find the speed of the projection and we know the magnitude of the velocity is equal to the speed of the projectile. So,

$s = 5\;{\rm{m/s}}$

**Thus, the speed of projection is 5 m/s.**

**Note:**Velocity is a vector quantity, it has both direction and magnitude while speed is a scalar quantity, and it has only magnitude. Velocity of a body can be equal to zero while the speed of the body cannot be equal to zero.

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