TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given, initial speed of particle, $v = (3 \hat{i}) \text{m/s}$ Acceleration of particle, $a = (-1 \hat{i} - 0.5 \hat{j}) \text{m/s}^2$ As we know that, $s_x = v_x t + \frac{1}{2} a_x t^2$. . . (i) and $s_y = v_y t + \frac{1}{2} a_y t^2$ . . . (ii) where, $v_x, v_y$ is the initial speed of particle along $X, Y$-axes respectively, $t$ be the time taken by the particle to reach maximum height. $\therefore \; s_x = 3t - \frac{1}{2}(1) t^2$ On differentiating both sides w.r.t $t$, we get $\frac{ds_x}{dt} = 3 - t = v_x'$ where, $v_x'$ is final speed of particle along $X$-axis. $\because$ At maximum distance\/ height, speed becomes $0 \text{ms}^{-1}$ $\Rightarrow \; t = 3 \text{s}$ $\therefore \; s_x = 3 \times 3 - \frac{1}{2}(1) \times 3^2$ $= 9 - \frac{9}{2} = \frac{9}{2} \text{m}$ $s_x = \frac{9}{2} \hat{i} \text{m}$ and $s_y = u_y t - \frac{1}{2} a_y t^2$ $\Rightarrow \; s_y = 0 \times 3 - \frac{1}{2} \times \frac{1}{2} \times 3^2$ $= \frac{-9}{4} \text{m}$ $\therefore \; s_y = -\frac{9}{4} \hat{j}$ $\because$ Particle reaches to maximum height and returns. $\therefore \; s_y = -\frac{9}{4} \times 2 \hat{j}$ $= -\frac{9}{2} \hat{j}$ $\therefore \; s = s_x \hat{i} + s_y \hat{j} = \frac{9}{2} \hat{i} - \frac{9}{2} \hat{j}$ $= \frac{9}{2} (\hat{i} - \hat{j})$