Note: This is a really simple beginner's physics question, I am just missing the connection.

The attachment says that when the projectile is launched from some height, the initial velocity is zero and the vertical displacement is negative the height (un-symmetrical projection). I don't understand why that is? How is the initial velocity of the projectile zero went the projectile is launched from some height.

If the projectile is not launched from ground level but from some height h, to find the time you need to use an equation that includes the vertical displacement. For "Plot on the graph's vertical axis," select y and then launch the projectile again. y (m) Graph: y vs. t Sketch the resulting graph. 60 1 The equation representing this vertical displacement is: 45- 30 - Ad, = v1,At-284t 15 - 0 t (s) (Again, a more specific version of one of our equations of motion.) 5 6 The same equation applies to a projectile launched horizontally from a height h. In that case, V1,=0 and the vertical displacement Ad,=-h . Rearranging for time, 2h At final = 1 for a horizontal launch from a height h g After you find the hang time, you can then find other quantities such as the range.