A projectile of mass 0.413 kg is shot from a cannon, at height 6.8 m, with an initial velocity vi having a horizontal component of 7.5m/s.

Question

The projectile rises to a maximum height of ∆y above the end of the cannon’s barrel and strikes the ground a horizontal distance ∆x past the end of the cannon’s barrel.
Determine the maximum height the projectile achieves after leaving the end of the cannon’s barrel.
Answer in units of m.

Answer 1

angle:
cosTheta=7.5/vi
time to fall 6.8m:
timein air=sqrt(6.5/g)
horizonal distance=7.5*timeinair
height:
mg*maxheight=1/2 m (vi*sinTheta)^2
solve for max height