A projectile of mass 0.695 kg is shot from a cannon, at height 6.5, with an initial velocity having a horizontal component of 5.8 m/s. The projectile rises to a maximum height of delta (y) above the end of the cannon's barrel and strikes the gound a horizontal distance delta (x) past the end of the cannon's barrel. g=9.8. The cannon is angled at 46 degrees. The height from the ground to delta y is 6.5 m. The maximum height delta y the projectile achieves after leaving the end of the cannon's barrel is 1.84m (which i calculated). I also calculated the vertical component of the initial velocity at the end of the cannon's barrel, where the projectile begins its trajectory is 6.01. I need to find the magnitude of the velocity vector when the projectile hits the ground? answer in units of m/s.

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vsage
this can pretty much be solved in two steps: one involves one of the four basic projectile motion equations. Which one do you think would be most appropriate for this problem?

v=vyo -2g(y-y0)

vsage
conkle08 said:
v=vyo -2g(y-y0)
That's the question I had in mind (using the information you gave) but I'm not sure that formula is correct as written, so i'll just give it here incase there is confusion: v^2 = vo^2 - 2g(y-y0)