Q 11-03-107JEE MainJEE Main 2022 (24 Jun, Shift 2)Medium
A body is projected from the ground at an angle of $45^\circ$ with the horizontal. Its velocity after $2\ \text{s}$ is $20\ \text{m s}^{-1}$. The maximum height reached by the body during its motion is ______ m. (use $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 20
At $45^\circ$, $u_x = u_y = a$. After 2 s: $v_x = a$, $v_y = a - 20$.
$$a^2 + (a-20)^2 = 400\ \Rightarrow\ 2a^2 - 40a = 0\ \Rightarrow\ a = 20\ \text{m s}^{-1}$$
$$H = \frac{u_y^2}{2g} = \frac{400}{20} = 20\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics