physics

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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?

  • physics -

    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

  • physics -

    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 = ?.

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