If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$, then $h$ is equal to $\frac{u^2}{2g}$:

Q.

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$, then $h$ is equal to $\frac{u^2}{2g}$. [$g$ is the acceleration due to gravity]

Q. If a body is thrown up with an initial velocity u and covers a height of h, then $h=\frac{u^2}{2g}$.

Q.

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$, then $h$ is equal to

Q. Question 2
A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) $\frac{u}{2}$
(b) $\frac{u^2}{2g}$
(c) $\frac{u^2}{g}$
(d) $\frac{u}{2g}$

Q. Assertion :Maximum possible height attained by a projectile is $\frac{u^2}{2g}$, where u is initial velocity Reason: To attain maximum height, a particle is thrown vertically upwards at $\theta ={90}^o$ to the ground.
$H=\frac{u^2}{2g}si{n}^2\theta \Rightarrow H_{max}=\frac{u^2}{2g}$