Question

The path is a projectile is a parabola.
A). True
B). False

Answer 1

Hint: Divide the initial velocity of the object into the horizontal component and vertical component. Then analyse each component separately to get the expression for the position of the object. Use the kinematic equations wherever required.

Formula used:
$\text{distance = speed }\!\!\times\!\!\text{ time}$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete step by step answer:
Suppose an object is projected from the ground with an initial velocity of $'u'm{{s}^{-1}}$at an angle $\theta $, at time t = 0. Since, the velocity of the object is a vector we can divide it into components and analyse the motion of the object for each component. Imagine a Cartesian plane in which the path of the projectile lies. Make then vertical direction as y-axis with the positive y-axis pointing upwards and the make horizontal direction as x-axis with the positive x-axis pointing towards the right. Place the origin of the plane at the point where the object is projected initially.

Divide the velocity ‘u’ into the components along the axis (x component and y component), as shown in the figure. Now we will analyse these two components separately.
Let us first analyse the x component i.e. the velocity along the positive x-axis direction. The value of the x component (ux) will be ${{u}_{x}}=u\cos \theta $. This velocity is along the horizontal direction where there is no force affecting its motion. Therefore, the object will not accelerate horizontally and the velocity will be constant with time. Let at some time t, the distance that the object in this direction be x. Using the formula $\text{distance = speed }\!\!\times\!\!\text{ time}$, we get, $x={{u}_{x}}t=u\cos (\theta )t$
$\Rightarrow t=\dfrac{x}{u\cos \theta }$ ……..(1)
Now, let us analyse the y component of the initial velocity (${{u}_{y}}=u\sin \theta $). In this direction, the motion of the object will be affected by the force of gravity. Since, the force of gravity is acting downwards, and the velocity is upwards, the object will decelerate. That means the acceleration of the object will be downwards i.e. $a=-g$. Let at some time t, the distance covered by the object in this direction be y. Now, use the formula $s=ut+\dfrac{1}{2}a{{t}^{2}}$.
Therefore, $y=u\sin (\theta )t+\dfrac{1}{2}(-g){{t}^{2}}$ ………(2)
Therefore, we found out how the position of the object changes with time. Now, substitute the value of t from equation (1) into equation (2)
We get,
$y=u\sin (\theta ).\left( \dfrac{x}{u\cos \theta } \right)+\dfrac{1}{2}(-g){{\left( \dfrac{x}{u\cos \theta } \right)}^{2}}$
$\Rightarrow y=\tan (\theta )x-\dfrac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }$
Make the coefficient of ${{x}^{2}}$ as 1 by multiplying both sides of the equation by $\left( -\dfrac{2{{u}^{2}}{{\cos }^{2}}\theta }{g} \right)$
$\Rightarrow {{x}^{2}}-\dfrac{2{{u}^{2}}\tan (\theta ){{\cos }^{2}}(\theta )}{g}x=-\dfrac{2{{u}^{2}}{{\cos }^{2}}\theta }{g}y$……. (3)
Since, u, g and $\theta $ are constant, let $\dfrac{2{{u}^{2}}\tan (\theta ){{\cos }^{2}}(\theta )}{g}$ be ‘k’ and $\dfrac{2{{u}^{2}}{{\cos }^{2}}\theta }{g}$ be ‘l’
Therefore, we can write the equation (3) as, ${{x}^{2}}-kx=-ly$
Use completing the square method on the left hand side of the equation.
 $\Rightarrow \left( {{x}^{2}}-kx+\dfrac{{{k}^{2}}}{4} \right)-\dfrac{{{k}^{2}}}{4}=-ly$
$\Rightarrow {{\left( x-\dfrac{k}{2} \right)}^{2}}=-ly+\dfrac{{{k}^{2}}}{4}$
$\Rightarrow {{\left( x-\dfrac{k}{2} \right)}^{2}}=-l\left( y-\dfrac{{{k}^{2}}}{4l} \right)$

The above equation is an equation of a parabola passing through the origin with $\left( \dfrac{k}{2},\dfrac{{{k}^{2}}}{4l} \right)$ as the centre. Hence, it is proved that the path taken by the projectile is a parabola.
Hence, the correct option is (A)true.

Note: We proved that the path of a projectile is a parabola. However, it is true only in ideal condition. In practical cases, this does not happen. In this solution, we have neglected the air resistance, which opposes the motion of an object. Hence, the projectile will not take a parabolic path. One more exception in this the solution is that we have taken the acceleration due to gravity (g) as constant. The value of g changes with altitude. For small heights, the change is not much and can be neglected.