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
Concept. At maximum height $v_y = \dfrac{dy}{dt} = 0$.
Step 1. Time. $5 - 2t = 0 \Rightarrow t = 2.5$ s.
Step 2. Height. $y = 12.5 - 6.25 = 6.25$ m.
Answer. $6.25$.
Solution by Sreeraj P, M.Sc Physics