Question

A projectile is given an initial velocity of (i^ + j^) . The cartesian equation of its path is....
$(a) \: y = ( x - {5x}^{2} )$
$(b) \: y = (2x - {5x}^{2} )$
$(c) (2x - {15x}^{2} )$
$(d) \: (2x - {25x}^{2} )$

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Answer 1

The standard Cartesian equation of the trajectory is,

$y=x\tan\theta-\dfrac {g}{2u^2\cos^2\theta}x^2$

where,

• $y$ = vertical displacement

• $x$ = horizontal displacement

• $\theta$ = angle of projection

• $u$ = initial velocity

The angle of projectile is equal to the angle made by the initial velocity with the horizontal. Thus,

$\tan\theta=\dfrac {1}{1}=1\quad\implies\quad\theta=45^{\circ}\\\\\therefore\ \cos^2\theta=\dfrac {1}{2}$

Also,

$u=\sqrt {1^2+1^2}=\sqrt 2$

Then,

$\dfrac {g}{2u^2\cos^2\theta}=\dfrac {10\cdot2}{2\cdot 2}=5$

Hence the equation of the trajectory is,

$\large\text { \underline {\underline {y=x-5x^2}} }$

Thus the right answer is (a).

Answer 2

Answer:

$\displaystyle{y=(x-5x^2)}$

a. option is correct.

Explanation:

Given :

Initial velocity u = $\displaystyle{\hat{i}+\hat{j}}$

Here :

$\displaystyle{u_x=1 \ \text{and} \ u_y=1$

Component of vectors :

u sin Ф = 1 and u cos Ф = 1

Now tan Ф = 1 / 1 ⇒ 1

We have equation of trajectory :

$\displaystyle{y=x.\tan\theta-\frac{g}{ 2u^2 \cos^2\theta} \ .x^2}$

Taking g = 10 and putting value of tan Ф

$\displaystyle{y=x\times1-\frac{10}{ 2\times1} \ .x^2}$

y = x - 5 x²

$\displaystyle{y=(x-5x^2)}$

Hence we get answer.