A stone mass M thrown straight up with initial velocity V reaches a height H. A second stone of mass 2M, thrown straight up with an initial velocity 2V. What will be its height?

Answer 1

for inital velocity v, we have

h(t) = vt-(g/2)t^2
This reaches a max height at t=v/g, and
h(v/g) = v(v/g) - (g/2)(v/g)^2
= v^2/g - v^2/2g = v^2/2g

So, we see that max height is directly proportional to v^2

So, double v, and you quadruple the max height.

M does not matter, as Galileo discovered.

Answer 2

according to the formula H=Vo*t-(gt^2)/2;
we can see that mass doesn't play any role here, so we can fint out our max. heights:
H2=2v*t2-(gt2^2)/2, and H1=v*t1-(gt1^2)/2;
Also t=v0/g => t1=v/g; t2=2v/g =>
H2=4v^2/g - 2v^2/g; H1=v^2/g-v^2/2g =>
H2=v^2(4/g-2/g); H1=v^2(1/g-1/2g) =>
H2/H1=4 => H2=4*H1. (I suggest to check my solution, mb I have mistakes)