physics

A projectile is fired at 30.0° above the horizontal. Its initial speed is equal to 172.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile? At what time after being fired does the projectile reach this maximum height?

asked by Kevin
  1. So in projectile problems horizontal a=0 so the velocity is constant. So finding x and y components for initial velocity
    X: 172.5m/s*cos30=149.4m/s
    Y: 172.5m/s*sin30=86.25m/s

    The max height is at the apex of the projectile, so use vertical velocity in
    Vfy =Viy+at---> -vi/a=t
    -86.25m/s)/-9.8m/s^2=8.80 sec

    Y=yo+vot+1/2at^2
    Y= 87.25m/s*8.8s +.5*9.8m/s^2*8.8s^2
    Y= 759m+379.456m
    Y= 1138m high

    Hope this helps!
    [vfy=0 at apex]

    posted by Kevin
  2. Vo = 172.5m/s[30o].
    Xo = 172.5*Cos30 = 149.4 m/s.
    Yo = 172.5*sin30 = 86.25 m/s.

    a. Yf^2 = Yo^2 + 2g*h.
    h = -(Yo)^2/2g = -(86.25)^2/19.6 = 379.5 m.

    b. Y = Yo + g*Tr = 0.
    Tr = -Yo/g = -86.25/-9.8 = 8.80 s. = Rise time.

    NOTE: In your Eq, use (-)9.8 for g.

    posted by Henry

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