Question

The height $y$ and horizontal distance $x$ covered by a projectile in a time $t$ seconds are given by the equations $y = \left( {8t - 5{t^2}} \right)$ and $x = 6t$ . If $x$ and $y$ are measured in meters, the velocity of projection is:
(A) $8m{s^{ - 1}}$
(B) $6m{s^{ - 1}}$
(C) $14m{s^{ - 1}}$
(D) $10m{s^{ - 1}}$

Answer 1

Hint : Use the expression of instantaneous velocity of a particle. The instantaneous velocity of a particle moving in $x$ direction is given by, $v = \dfrac{{dx}}{{dt}}$ , where $v$ is the velocity of the particle at an instant of time .

Complete Step By Step Answer:
We know, The instantaneous velocity of a particle is given by, $\vec v = \dfrac{{d\vec s}}{{dt}}$ where $\vec v$ is the velocity vector of the body and $\vec s$ is the displacement vector.
If we write this equation in $2D$ Cartesian coordinate that becomes, ${v_x}\hat i + {v_y}\hat j = \dfrac{d}{{dt}}(x\hat i + y\hat j)$ where ${v_x},{v_y}$ are the velocity components and $x,y$ are displacement or position component along $x$ and $y$ . Comparing this we can write, ${v_x} = \dfrac{{dx}}{{dt}}$ and ${v_y} = \dfrac{{dy}}{{dt}}$
So, we have given here, $y = \left( {8t - 5{t^2}} \right)m$ and $x = 6tm$ .
Therefore, differentiating these two equation with respect to $t$ we get the velocity component of $x,y$ coordinate in any instant of time as,
${v_x} = \dfrac{d}{{dt}}(6t)$
$\Rightarrow {v_x} = 6$ ,Which is independent of time.
And, ${v_y} = \dfrac{d}{{dt}}\left( {8t - 5{t^2}} \right)$
$\Rightarrow {v_y} = 8 - 5 \cdot 2t$ Where ${v_y}$ is the velocity along $y$ component.
$\Rightarrow {v_y} = 8 - 10t$
Now, we have to find the initial velocity of the particle is given at $t = 0$ .
Therefore, initial velocity along $y$ component becomes, ${v_y} = 8 - 10 \cdot 0 = 8$
initial velocity along $x$ component becomes,
${v_x} = 6$ at $t = 0$
hence, velocity of the projectile can also be written as, $\vec v = {v_x}\hat i + {v_y}\hat j$ at initial condition $t = 0$
Putting the, the values of ${v_x}$ , ${v_y}$ , we can get, $\vec v = 8\hat i + 6\hat j$
If we take the magnitude of the velocity $\vec v$ that becomes,
$\left| {\vec v} \right| = \sqrt {{8^2} + {6^2}} m{s^{ - 1}}$
On simplifying then we get, $\left| {\vec v} \right| = 10m{s^{ - 1}}$
Hence, the velocity of projection of the particle is $10m{s^{ - 1}}$ .
Hence, option ( D) is correct.

Note :
For a particle projected under the influence of a gravitational field the velocity along horizontal is always constant only the vertical component of velocity changes due to the gravity. one can always check this condition to find whether the motion is a projectile motion or not. Projectile motion is always constrained in a plane, it is a two dimensional motion. Note that when there is an air blow or any force along the other axis is present then motion will be a three dimensional motion.