Q 11-03-043JEE MainMedium
The position of a particle moving in a vertical plane is $\vec{r} = \left[2t\,\hat{i} + (5t - t^2)\,\hat{j}\right]$ m, where $\hat{j}$ is vertically upward and $t$ is in seconds. Find the maximum height ($y$-coordinate) reached by the particle, in metres.
Numerical value type. Enter your answer.
Answer: 6.25
$$v_y = \frac{dy}{dt} = 5 - 2t = 0 \;\Rightarrow\; t = 2.5\ \text{s}$$
$$y_{max} = 5(2.5) - (2.5)^2 = 12.5 - 6.25 = 6.25\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics