Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. Assume free fall. Part A If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return

36s

Explanation:

Let the objects be A and B.

Let the initial velocity of A be U and the initial velocity of B be 3U

The height sustain by A will be;

The final velocity would be zero

V2 = U2-2gH

Hence

0^2= U2 -2gH

H = U^2/2g

Similarly for object B, the height sustain is;

V2 = (3U)^2-2gH

Hence

0^2= 3U^2 -2gH

U2-2gH

Hence

0^2= U2 -2gH

H = 3U^2/2g

By comparism. The object with higher velocity sustains more height and so should fall longer than object A.

Now object A would take;

From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;

V=10×12=120m/s let g be 10m/S2

Similarly for object B,

The final velocity for B when it's falling it should be 3×that of A

Meaning

3V= gt

t =3V/g = 3× 120/10 = 36s

double the distance = divide strength by 4