JEE Main · 2013 · 07 Apr 2013 · Offline · Kinematics

A projectile is given an initial velocity of

(\hat{i} + 2 \hat{j}) m / s

, where

\hat{i}

is along the ground and

\hat{j}

is along the vertical. If

g = 10 m / s^{2}

, the equation of its trajectory is:

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

Answer: (A)

Solution

The launch velocity splits into

u_{x} = 1 m/s

along the ground and

u_{y} = 2 m/s

upward. Horizontally the motion is uniform,

x = u_{x} t = t

, so time and horizontal distance are numerically the same. Vertically,

y = u_{y} t - \frac{1}{2} g t^{2} = 2 t - 5 t^{2}

with

g = 10 m/s^{2}

. Replacing

t

x

gives the trajectory

y = 2 x - 5 x^{2}

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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