A projectile is given an initial velocity of i +2 j . The cartesian equation of its path is (Take g=10 ~m ~s^ -2 )

Correct answer: B. y=2 x-5 x^2 Given, u = i +2 j =u_x i +u_y j Then u_x=1=u θ and u_y=2=u θ θ=(u θ)/(u θ)=(2)/(1)=2 The equation of trajectory of a projectile motion is aligned & y=x θ-(g x^2)/(2 u^2 ^2 θ)=x θ-(g x^2)/(2(u θ)^2) & y=x × 2-(10 × x^2)/(2(1)^2)=2 x-5 x^2 aligned

AIIMS 2017

Exam: AIIMS
Year: 2017
Subject: Physics
Chapter: Motion in Two Dimensions

AIIMS 2017 · Physics · Motion in Two Dimensions

A projectile is given an initial velocity of i +2 j . The cartesian...

A projectile is given an initial velocity of

\hat{i}+2 \hat{j}

. The cartesian equation of its path is (Take

g=10 \mathrm{~m} \mathrm{~s}^{-2}

)

Options

A $y=x-5 x^2$
B $y=2 x-5 x^2$
C $y=2 x-15 x^2$
D $y=2 x-25 x^2$

Correct answer

B. $y=2 x-5 x^2$

Step-by-step solution

Given,

\vec{u}=\hat{i}+2 \hat{j}=u_x \hat{i}+u_y \hat{j}

Then

u_x=1=u \cos \theta

and

u_y=2=u \sin \theta \therefore \quad \tan \theta=\frac{u \sin \theta}{u \cos \theta}=\frac{2}{1}=2

The equation of trajectory of a projectile motion is

\begin{aligned} & y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}=x \tan \theta-\frac{g x^2}{2(u \cos \theta)^2} \\ & \therefore \quad y=x \times 2-\frac{10 \times x^2}{2(1)^2}=2 x-5 x^2\end{aligned}

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