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(A) The initial velocity of the projectile is $\vec{u} = (1\hat{i} + 2\hat{j}) 	ext{ ms}^{-1}$.Thus,the initial horizontal component is $u_x = 1 	ex

Vedclass · Accepted Answer

$y=2x-5x^2$

Vedclass · Answer

(A) The initial velocity of the projectile is $\vec{u} = (1\hat{i} + 2\hat{j}) 	ext{ ms}^{-1}$.Thus,the initial horizontal component is $u_x = 1 	ext{ ms}^{-1}$ and the initial vertical component is $u_y = 2 	ext{ ms}^{-1}$.The acceleration components are $a_x = 0$ and $a_y = -g = -10 	ext{ ms}^{-2}$.At any time $t$,the horizontal distance covered is:$x = u_x t = 1 \cdot t \implies t = x \quad \dots (i)$The vertical distance covered is:$y = u_y t + \frac{1}{2} a_y t^2$$y = 2t + \frac{1}{2}(-10)t^2$$y = 2t - 5t^2 \quad \dots (ii)$Substituting the value of $t$ from equation $(i)$ into equation $(ii)$,we get:$y = 2(x) - 5(x)^2$$y = 2x - 5x^2$Therefore,the correct option is $A$.

$A$ projectile is given an initial velocity of $(\hat{i}+2 \hat{j}) \text{ ms}^{-1}$. The equation of its path is $(g=10 \text{ ms}^{-2})$

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