Physics ( Equations of motion ) question?

1st ball :
Initial speed upward = U
Final speed at maximum height H = 0
Hence, 0 = U² - 2gH [ applying the equation V² = U² + 2as , V = 0 , a = - g and s = H ]
=> H = U² / 2g .......... (1)
2nd ball :
Initial speed upward = 2U (as per the question)
Final speed at the maximum height H' = 0 [ applying the same equation as above, V = 0 , a = - g
and s = H' ]
Hence, 0 = (2U)² - 2gH' = 4U² - 2gH'
=> H' = (4U²) / (2g) = 2U² / g ........ (2)
From (1) and (2), H / H' = {U² / (2g)} / {(2U²) / g} = 1 / 4

Energy = 1/2 m v^2
and energy = mgh
so h = v^2 / 2g

if v is doubled then v^2 increases by a factor of 4
so the height also increases by a factor of 4.

We have H = Uy^2/2g = (2uy)^2/2g and h = uy^2/2g; so that H/h = 4uy^2/2g//uy^2/2g = 4. The H = 4h so that the height for the faster toss Uy = 2 uy will be four times higher. ANS.

formula for Hmax is given by
Hmax = V^2[sin^2(x)]/(2g)
so if V1:V2=1:2
and angle of projection x is same for both balls that is 90 so
Hmax is given by
(1^2)/(2^2)=1/4
simple......