Urgent physics question?

The impact speed would NOT double. Neither in an ideal world nor in a practical one.

V^2 = u^2 + 2as

where V is the final speed.

let the two initial speeds be u1 and u2 such that u2 = 2 u1

V2^2/V1^2 = ((2u1)^2 + 2as) / ( u1^2 + 2as)

(v2/v1) = sqrt( ((2u1)^2 + 2as) / ( u1^2 + 2as) )

Now you can do the algebraic expansion to find this ratio for any height but by substituting values you find that

V2/v1= sqrt( (2u1)^2 / u1^2)

= sqrt( 4 u1^2/u1^2) = sqrt(4) = 2 ONLY if either a is zero ( no gravity ) Or s is zero ( no height)

At all other conditions the velocity v2 is always less than 2 v1

to demonstrate (2a+ b) / ( a+b) = ( ( 2a+2b) - b) / ( a+b) = 2 - b/(a+b)

ie the result is always less than 2.

Now if it is going FASTER there is no possibility that the air time would INCREASE.

That is quite absurd.

However if the final speed is LESS than double, then the resulting air time would be GREATER than one half of the original air time.

Let's have a numerical exemple :

assume h = 50 m , V1 = 10 m/sec , V2 20 m/sec and g = 10m/sec^2

V1 case

-50 = -10*t1-g/2*t1^2

falling time t1 = (10-√10^2+5*4*50)/-10 = 2.32 sec

impact speed Vi1 = √V1^2+2gh = √10^2+2*10*50 = 33.2 m/sec

V2 case

-50 = -20*t1-g/2*t1^2

falling time t2 = (20-√20^2+5*4*50)/-10 = 1.74 sec

impact speed Vi2 = √V2^2+2gh = √20^2+2*10*50 = 37.4 m/sec

t1/t2 = 1.33

Vi2/Vi1 = 1.13