Answer
8.3 kilograms
Work Step by Step
We know that the first projectile will have gone half of its path, which is:
$d = \frac{(380)^2sin(2\times55)}{2\times 9.81}=6915 \ m$
The height will be:
$h = \frac{(380)^2sin^2(55)}{2\times 9.81}=4939 \ m$
We know that its velocity will be:
$v=380cos(55)=218\ m/s$
We now find how long it will take the two projectiles to land:
$t = \sqrt{\frac{2(4939)}{g}}=31.7 \ s$
Using this, we find the initial velocity:
$v_0=\frac{-6915+9600}{31.7}=84.7 \ m/s$
Since momentum is conserved, we find:
$(84.7)(14+m)=218(14)+140m$
$m=8.3 \ kg$
