Question

TThe equation of motion of a projectile is $y=ax-b{{x}^{2}}$ where a and b are constants of motion. Match the quantities in Column I with the relations in Column II.

Column I	Column II
(A) The initial velocity of projectile	(p) $\dfrac{a}{b}$
(B) The horizontal range of projectile	(q) $a\sqrt{\dfrac{2}{bg}}$
(C) The maximum vertical height attained by projectile	(r) $\dfrac{{{a}^{2}}}{4b}$
(D) The time of flight of projectile	(s) $\sqrt{\left( {{a}^{2}}+1 \right)\dfrac{g}{2b}}$

A. A-p, B-q, C-r, D-s
B. A-s, B-p, C-q, D-r
C. A-s, B-p, C-r, D-q
D. A-p, B-s, C-r, D-q

Answer 1

Hint: Differentiate the given equation of motion with respect to time and find an equation involving the horizontal and vertical velocities. Again differentiate the equation with respect to time. Here the vertical acceleration is –g and horizontal acceleration is zero. With these equations find the initial velocity. Horizontal range is when x=0. Maximum height is attained when vertical velocity is zero. For time of flight d=vt.

Formula used:
d=vt

Complete step by step answer:
Let play with the given equation of motion of the projectile and then see whether we get something useful.
Consider the equation of motion $y=ax-b{{x}^{2}}$ ……. (i).
Let us assume that the point where the projectile is launched be the origin and the y and x coordinates are the displacements of the projectile in the horizontal and vertical directions respectively.
First, differentiate equation (i) with respect to time t.
Hence, we get,
$\dfrac{dy}{dt}=a\dfrac{dx}{dt}-2bx\dfrac{dx}{dt}$ ……. (ii).
We know that $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ are the horizontal and vertical velocities respectively.
Let $\dfrac{dx}{dt}={{v}_{x}}$ and $\dfrac{dy}{dt}={{v}_{y}}$
Hence, equation (ii) can be written as,
${{v}_{y}}=a{{v}_{x}}-2bx{{v}_{x}}$ …… (iii).
Now differentiate equation (iii) with respect to time to find the horizontal and vertical accelerations of the projectile.
After differentiating w.r.t time we get,
$\dfrac{d{{v}_{y}}}{dt}=a\dfrac{d{{v}_{x}}}{dt}-2b\dfrac{dx}{dt}{{v}_{x}}-2bx\dfrac{d{{v}_{x}}}{dt}$
Let $\dfrac{d{{v}_{y}}}{dt}={{a}_{y}}$ and $\dfrac{d{{v}_{x}}}{dt}={{a}_{x}}$.
Hence, we get,
${{a}_{y}}=a{{a}_{x}}-2b{{v}_{x}}{{v}_{x}}-2bx{{a}_{x}}$
$\Rightarrow {{a}_{y}}=a{{a}_{x}}-2bv_{x}^{2}-2bx{{a}_{x}}$ ….. (iv).
However, during a projectile motion, its horizontal acceleration is zero and the vertical acceleration is equal to –g.
Therefore, ${{a}_{x}}=0$ and ${{a}_{y}}=-g$
Substitute the above values in equation (iv).
$\Rightarrow -g=a(0)-2bv_{x}^{2}-2bx(0)$
$\Rightarrow g=2bv_{x}^{2}$
$\Rightarrow {{v}_{x}}=\sqrt{\dfrac{g}{2b}}$ ….. (v).
Since the horizontal acceleration is zero, the horizontal velocity is always constant i.e. ${{v}_{x}}=\sqrt{\dfrac{g}{2b}}$.
Now, let us find the initial velocity of the projectile. Let the initial velocity be u.
We know that the initial velocity is at the origin (x=0,y=0) and $v_{x}^{2}+v_{y}^{2}={{u}^{2}}$ ….. (vi).
Substitute x=0 and ${{v}_{x}}=\sqrt{\dfrac{g}{2b}}$ in equation (iii).
${{v}_{y}}=a\sqrt{\dfrac{g}{2b}}-2b(0){{v}_{x}}$
$\Rightarrow {{v}_{y}}=a\sqrt{\dfrac{g}{2b}}$
Substitute ${{v}_{y}}=a\sqrt{\dfrac{g}{2b}}$ and ${{v}_{x}}=\sqrt{\dfrac{g}{2b}}$ in equation (vi).
${{\left( a\sqrt{\dfrac{g}{2b}} \right)}^{2}}+{{\left( \sqrt{\dfrac{g}{2b}} \right)}^{2}}={{u}^{2}}$
$\Rightarrow {{a}^{2}}\dfrac{g}{2b}+\dfrac{g}{2b}={{u}^{2}}$
$\Rightarrow \left( {{a}^{2}}+1 \right)\dfrac{g}{2b}={{u}^{2}}$
$\Rightarrow u=\sqrt{\left( {{a}^{2}}+1 \right)\dfrac{g}{2b}}$.
Therefore, the matching option for A is s.
Let's find the horizontal range of the projectile.
Horizontal range is the maximum horizontal displacement of the projectile from the point of release (i.e. origin).
The maximum horizontal displacement (R) lies on the x-axis. Here, y=0 and x=R.
Therefore, substitute y=0 and x=R in equation (i).
$\Rightarrow 0=aR-b{{R}^{2}}$
$\Rightarrow 0=R(a-bR)$
This means either R=0 or (a-bR)=0.
R=0 cannot be the horizontal range and hence this value is discarded.
This implies that (a-bR)=0.
$\Rightarrow R=\dfrac{a}{b}$.
Therefore, the matching option for B is a.
Let us find the maximum vertical height (H) attained by the projectile.
We know that at the maximum height ${{v}_{y}}=0$. Substitute this value in equation (iii).
$\Rightarrow 0=a{{v}_{x}}-2bx{{v}_{x}}$
$\Rightarrow 0=a-2bx$
$\Rightarrow x=\dfrac{a}{2b}$
Substitute $x=\dfrac{a}{2b}$ and y=H in equation (i).
$H=a\left( \dfrac{a}{2b} \right)-b{{\left( \dfrac{a}{2b} \right)}^{2}}$
$H=\dfrac{{{a}^{2}}}{2b}-\dfrac{{{a}^{2}}}{4b}=\dfrac{{{a}^{2}}}{4b}$
Therefore, the matching option for C is r.
Let us find the time flight of projectile (T). It is the time for which the projectile is in air.
We already know that ${{v}_{x}}=\sqrt{\dfrac{g}{2b}}$ and is constant.
Therefore, use the formula $x={{v}_{x}}t$.
$\Rightarrow x=\left( \sqrt{\dfrac{g}{2b}} \right)t$ ….. (vii).
When the time is equal to T, $x=R=\dfrac{a}{b}$.
Substitute t=T and $x=\dfrac{a}{b}$ in equation (vii).
$\Rightarrow \dfrac{a}{b}=\left( \sqrt{\dfrac{g}{2b}} \right)T$
$\Rightarrow T=\dfrac{a}{b}\sqrt{\dfrac{2b}{g}}=a\sqrt{\dfrac{2}{bg}}$
Therefore, the matching option for D is q.
Hence, the correct option is C. A-s, B-p, C-r, D-q.

Note: The maximum height can also be found if you know that the parabolic path of the projectile is symmetric about the vertical line passing through the point of maximum height. Therefore, when the projectile is at maximum vertical height, $x=\dfrac{R}{2}$. Hence, you can directly substitute $x=\dfrac{R}{2}$ in equation (i).