Projectile Range Comparison

## Question:

Can you answer this question?

Two projectiles are launched from ground level at the same angle
above the horizontal, and both return to ground level. Projectile
A has a launch speed that of projectile B. Assuming that air
resistance is absent, what should be the ratio of the ranges?
## Answer:

I think your question is missing some data. If you intend that
projectile A has twice the launch velocity as B but otherwise
identical, you might think of it like this. The horizontal and
vertical component of the faster projectile are both doubled ( in
the scenario I have set up). Assuming that neither go far enough
that the Earth's curvature has to be considered, the height
to which each projectile will rise is
m*g*h=1/2*m*vh^{2} - conservation of energy. Since
vhA is twice vhB, vhA^{2} is 4 times vhB^{2} so
hA=4*hB
The time required for A to come down is given by 4*h=1/2*a*tA^{2}. The time for the B to
fall from its peak is given by
h=1/2*a*tB^{2}. Dividing these equations we get
4=tA^{2}/tB^{2}. Taking
the square root we get 2=tA/tB or tA=2*tB. Of course the rise time will equal the
fall time so the relationship of total times in the air will be
the same. Since A is in the air 2 times as long as B and its
horizontal velocity is twice as much, A will go 4 times as far as
B.