If you shoot a cannonball horizontally and at its apex of 18 meters it is traveling 77 m/s, what was its muzzle velocity? (It weighs 4.2 kg). Thanks!
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If it is shot horizontally it won't rise to a height of 18m.
So I'll ignore the horizontal bit.
Let Initial vertical velocity component be Vyi
so Viy^2 = 2gh = 2*9.8*18 = 352.8 m^2/s^2
Viy = sqrt(352.8) = 18.8 m/s
At the apex, Vy = 0, so v = Vx = 77 m/s
So the muzzle velocity was
v = sqrt(Vx^2 + Viy^2) = 79.3 m/s
(and the "horizontal" angle was arctan(18.8/77) = 13.7 deg :)
So I'll ignore the horizontal bit.
Let Initial vertical velocity component be Vyi
so Viy^2 = 2gh = 2*9.8*18 = 352.8 m^2/s^2
Viy = sqrt(352.8) = 18.8 m/s
At the apex, Vy = 0, so v = Vx = 77 m/s
So the muzzle velocity was
v = sqrt(Vx^2 + Viy^2) = 79.3 m/s
(and the "horizontal" angle was arctan(18.8/77) = 13.7 deg :)