Q 11-03-001JEE MainTop questionEasy
A ball is projected with speed $20\ \text{m/s}$ at $30^\circ$ above the horizontal. Taking $g = 10\ \text{m/s}^2$, the maximum height reached is
Answer: (A) $5\ \text{m}$
Only the vertical component of velocity matters for height: $u_y = u\sin 30^\circ = 10\ \text{m/s}$.
$$H = \frac{u_y^2}{2g} = \frac{10^2}{2\times 10} = 5\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics