Q78 / 90 · JEE Main 2013 (07 Apr) · Physics

A projectile is given an initial velocity of $(\hat{i} + 2 \hat{j}) m s^{- 1}$ , where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical upward. If $g = 10 m s^{- 2}$ , the equation of its trajectory is :

A
$4 y = 2 x - 5 x^{2}$
B
$4 y = 2 x - 25 x^{2}$
C
$y = x - 5 x^{2}$
Correct answer
D
$y = 2 x - 5 x^{2}$

Solution

\vec{v} = \hat{i} + 2 \hat{j}

\Rightarrow x = t

..... (i)

y = 2 t - \frac{1}{2} (10 t^{2})

..... (ii)

From (i) and (ii)

y = 2 x - 5 x^{2}

Hence, the correct answer is option: 'D'.

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