# physics

A cannonball is shot (from ground level) with an initial horizontal velocity of 41 m/s and an initial vertical velocity of 22 m/s.

1) What is the speed of the cannonball 1.7 seconds after it was shot?

2) How high above the ground is the cannonball 1.7 seconds after it is shot?

1. 1) the horizontal velocity stays constant
... the vertical velocity is decreased by gravitational acceleration ... 9.8 m/s^2
... 22 - (1.7 * 9.8)
... add the two velocity vectors to find the resultant

2) you have the vertical velocity at 1.7 s
... average it with the initial (22 m/s)
... multiply the average by 1.7 s to find the height

posted by Scott
2. Vo = Xo + Yo = 41 + 22i m/s.

1. V = Y = Yo + g+t = 22 - 9.8*1.7 = 5.34 m/s.
Vo = Xo + Yi = 41 + 5.34i = (Convert to polar form).

2. Y^2 = Yo^2 + 2g*h.
5.34^2 = 22^2 - 19.6h, h = ?.

posted by Henry

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