## Essential University Physics: Volume 1 (3rd Edition)

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: $(84.7)(14+m)=218(14)+140m$ $m=8.3 \ kg$