Question

A particle is thrown vertically upwards. Its velocity at half of the maximum height is $20\text{m/s}$. Then the maximum height attained by it will be: (Take $g=10{\text{m/s}}^2$)

• A. $40\text{m}$
• B. $25\text{m}$
• C. $10\text{m}$
• D. $45\text{m}$

Answer 1

The correct option is A $40\text{m}$

Let the maximum height attained by particle be $H$
Taking vertically upward direction as $+ve$ and applying kinematic equation,
$v^2=u^2+2as...\left(i\right)$

where at maximum height $v=0$,
Displacement of particle from half the maximum height to maximum height is $s=\frac{+H}{2}$
& velocity of particle at distance equal to half of maximum height is $u=+20\text{m/s}$
Also, $a=-g=-10{\text{m/s}}^2$

Putting all values in Eq $\left(i\right)$,
$(0{)}^2=(20{)}^2-2\times 10\times \left(\frac{H}{2}\right)$
or $400=10H$
$\therefore H=40\text{m}$