An object thrown vertically upward reaches a height of 500 m. Taking $g = 10$ m/s$^2$, its initial velocity was:

1. At maximum height $v = 0$ and acceleration $a = -g = -10$ m/s$^2$. 2. Use $v^2 = u^2 + 2as$: $0 = u^2 + 2(-10)(500)$. 3. So $u^2 = 2 \times 10 \times 500 = 10000$. 4. $u = \sqrt{10000} = 100$ m/s; halving to 50 m/s ignores the factor 2 and is the trap. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 1 "Gravitation", p.24_