Kinematics Ques 50
- A projectile is given an initial velocity of $(\mathbf{i}+2 \mathbf{j}) m / s$, where, $\mathbf{i}$ is along the ground and $\mathbf{j}$ is along the vertical. If $g=10 m / s^{2}$, then the equation of its trajectory is
(2013 Main)
(a) $y=x-5 x^{2}$
(b) $y=2 x-5 x^{2}$
(c) $4 y=2 x-5 x^{2}$
(d) $4 y=2 x-25 x^{2}$
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Answer:
Correct Answer: 50.(b)
Solution:
Formula:
- Initial velocity $=(\mathbf{i}+2 \mathbf{j}) m / s$
Magnitude of initial velocity, $u=\sqrt{(1)^{2}+(2)^{2}}=\sqrt{5} m / s$
Equation of trajectory of projectile is
$$ \begin{aligned} y & =x \tan \theta-\frac{g x^{2}}{2 u^{2}}\left(1+\tan ^{2} \theta\right) \quad \Big[\tan \theta=\frac{y}{x}=\frac{2}{1}=2\Big] \\ \therefore \quad y & =x \times 2-\frac{10(x)^{2}}{2(\sqrt{5})^{2}}\left[1+(2)^{2}\right] \\ & =2 x-\frac{10\left(x^{2}\right)}{2 \times 5}(1+4) \\ & =2 x-5 x^{2} \end{aligned} $$
BETA
