An object is thrown vertically upwards with a velocity u, the greatest height h to which it will rise before falling back is given by:
(a) $u/g$
(b) $u^2/2g$
(c) $u^2/g$
(d) $u/2g$

Here, initial velocity $=u$

Maximum height $=h$

At reaching the maximum height, the final velocity $v=0$

On substituting the above values in the equation of motion,

$v^2=u^2+2gh$

$0^2=u^2+2(-g)h$     [The motion of the object is vertically upward, so

$g$ will be negative]

Or $0=u^2-2gh$

Or $u^2=2gh$

Or $h=\frac{u^2}{2g}$

So, option (b) is correct.

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Updated on: 10-Oct-2022

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