# physics

A shell leaves a mortar with a muzzle velocity of 500 ft/s directed upward at 60 degrees with the horizontal. Determine the position of the shell and its resultant velocity 20 s after firing. how high will it rise?

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1. Vo = 500Ft/s[60o].
Xo = 500*Cos60 = 250 Ft/s.
Yo = 500*sin60 = 433 Ft/s.

Y = Yo + g*Tr = 0, 433 - 32Tr = 0, Tr = 13.53 s. Rise time.

h = Yo*t + 0.5g*t^2.
h = 433*20 - 16*20^2 = 2260 Ft.
and falling.

Y = Yo + g*t = 433 - 32*20 = -207 Ft/s = 207 Ft/s, Downward.

V = Xo + Yi = 250 - .53= 325 Ft/s[-39.6o].

h max = Yo*Tr + 0.5g*Tr^2.
h max = 433*13.53 - 16*13.53^2 = 2930 Ft.

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2. Correction: V = Xo + Yi = 250 - 207i = 325Ft/s[-39.6o].

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