Question

A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then maximum height attained by it is (Take $g = 10$ m/s²)

• A. 8 m
• B. 10 m
• C. 12 m
• D. 16 m

Answer 1

Answer: (B)

10 m

Answer 2

Let particle thrown with velocity u and its maximum height is $H$ then $H = \frac{u^{2}}{2g}$

When particle is at a height $H/2$, then its speed is $10m/s$

From equation $v^{2} = u^{2} - 2gh$,

$(10)^{2} = u^{2} - 2g\left( \frac{H}{2} \right) = u^{2} - 2g\frac{u^{2}}{4g}$

⇒ $u^{2} = 200$

$\therefore$ Maximum height $H = \frac{u^{2}}{2g} = \frac{200}{2 \times 10}$= 10m