# The Point from where a ball is projected is taken as the origin of the coordinate axes. The $x$ and $y$ components of its displacement are given by $x = 6t$ and $y = 8t - 5{t^2}$. What is the velocity of projection?

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**Hint:**In order to find the magnitude of velocity we will use the derivative of displacement in both the components in given $x$ and $y$ direction and then using vector algebra we will find net magnitude of the velocity and its direction.

**Complete step by step answer:**

As we know that velocity and displacement are related as ${v_x} = \dfrac{{dx}}{{dt}}$ in the x direction and in y direction it can be calculated as ${v_y} = \dfrac{{dy}}{{dt}}$ .

we have given that, $x = 6t$ taking derivative of this we will get,

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

$\Rightarrow {v_x} = 6\hat{ i}$

Now, we will find the velocity in y direction using $y = 8t - 5{t^2}$ we will get,

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

$\Rightarrow {v_y} = 8 - 10t$

In origin we have $t = 0$ so we get,

${v_y} = 8\hat j$

Hence, net velocity can be written together in the vector form as:

$\vec v = 6\hat i + 8\hat j$

Now, the resultant velocity of the projectile can be found using the formula we have,

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

On putting the magnitudes we get,

$\left| v \right| = \sqrt {64 + 36} $

$\therefore \left| v \right| = 10\,m\,{\sec ^{ - 1}}$

**Hence, the magnitude of velocity of the projectile is $\left| v \right| = 10\,m\,{\sec ^{ - 1}}$ and its direction of velocity of the projectile can be written as $\vec v = 6\hat i + 8\hat j$.**

**Note:**It should be remembered that, the basic formulas of derivation of one variable with respect to other with functions like $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$ and the resultant magnitude of a given vector in two dimensional form by calculated as $R = \sqrt {{P^2} + {Q^2}} $ where $P$ and $Q$ are the two components of a vector $R$ in $x$ and $y$ directions respectively.

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