Question

A particle is projected from the ground with an initial velocity of $20{m}/{s}\;$ at an angle of ${{30}^{\circ }}$ with horizontal. The magnitude of change in velocity in a time interval from $t=0$ to $t=0.5s$ is $\left( g=10{m}/{{{s}^{2}}}\; \right)$.
$\begin{aligned} & \left( A \right)5{m}/{s}\; \\\ & \left( B \right)2.5{m}/{s}\; \\\ & \left( C \right)2{m}/{s}\; \\\ & \left( D \right)4{m}/{s}\; \\\ \end{aligned}$

Answer 1

Projectile motion is that the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The thing is named a projectile, and its path is named its trajectory. So, firstly calculate the initial and final velocity obtained by the particle at a given time and then find the difference between them.

Formula used:
For initial velocity:
$u=ucos\theta \hat{i}+usin\theta \hat{j}$
For final velocity:
$\vec{v}=\vec{u}+\vec{a}t$

Complete answer:
As given in problem:

& u=20m/s~ \\\ & \theta ={{30}^{\circ }}~ \\\ & ~t=0.5s \\\ \end{aligned}$$ So, initial velocity: $\begin{aligned} & \vec{u}=ucos\theta \hat{i}+usin\theta \hat{j} \\\ & \vec{u}=u(cos{{30}^{\circ }}\hat{i}+sin{{30}^{\circ }}\hat{j}) \\\ & \vec{u}=20(\dfrac{\sqrt{3}}{2}\hat{i}+\dfrac{1}{2}\hat{j}) \\\ & \vec{u}=10\sqrt{3}\hat{i}+10\hat{j} \\\ \end{aligned}$ Acceleration $a=-g\hat{j}=-10\hat{j}$ (acceleration is acting downward) $\vec{v}=\vec{u}+\vec{a}t$ $\vec{v}=\vec{u}+\vec{a}t$ $\vec{v}=(10\sqrt{3}\hat{i}+10\hat{j})-(10\hat{j}\times 0.5)$ $v=10\sqrt{3}\hat{i}+5\hat{j}$ Now, change in velocity = $\Delta \vec{v}=\vec{v}-\vec{u}=-5\hat{j}$ So, the magnitude of the change in velocity = $5{m}/{s}\;$ The correct option is $5{m}/{s}\;$ **So, the correct answer is “Option A”.** **Additional Information:** When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that's directed towards the middle of the world (we assume that the particle remains on the brink of the surface of the earth). The trail of such a particle is named a projectile and therefore the motion is named projectile motion. Air resistance to the motion of the body is to be assumed absent in projectile motion. In a Projectile Motion, there are two independent rectilinear motions: Along the x-axis: uniform velocity, liable for the horizontal (forward) motion of the particle. Along y-axis: uniform acceleration, liable for the vertical (downwards) motion of the particle. **Note:** A short method is- $$\Delta v=a\times t=10\times 0.5=5m/s$$ Use the proper sign for velocities and acceleration when the particle is moving on its trajectory and the sign conventions are according to the acceleration due to gravity. If you are using acceleration due to gravity as positive downward then rest of the signs should take according to it and vice-versa.