A body is throw up with a velocity ‘u’. It reaches maximum height ‘h’. If its velocity of projection is doubled the maximum height it reaches is ___

initial velocity = u

maximum height = h

final velocity = v

we know,

v² - u² = 2as

s = v² - u² / 2a

h = -u² / - 2g    {  v = 0 (as it reaches maximum height ) }

h = u² / 2g

Now,

If velocity is doubled

i.e u = 2u

u² = (2u)² = 4u²

h = 4u² / 2g

h = 4 h

So, The maximum height is 4 times the initial height.