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Projectile Motion Help, Please?

  1. Oct 12, 2003 #1
    Well, I'm not sure how to end this problem. Is it correct how I approached it?

    A basketball leaves a player's hands at a height of 2.1 m above the floor. The basket is 2.6 m above the floor. The player likes to shoot the ball at a 35* (degree) angle. If the shot is made from a horizontal distance of 12.0 m and must be accurate to +or- 0.22 m (horizontally), what is the range of initial speeds allowed to make the basket?

    vx = v1cos35*
    vy = v1sin35*

    t = dx/v1cos35*
    dx = 12.0 m
    d = 0.5 m
    d = v1t + 1/2gt^2

    0.5 m = v1sin35* x (12.0m/v1cos35*) + 1/2(-9.8m/s^2)(12.0m/v1cos35*)^2
    All cleaned up all that work and got v1 = 11 m/s.

    dx = 12.0m + 0.22m = 12.22 m
    dx = 12.0m - 0.22m = 11.78 m

    t = dx/vx
    t = 12.0m/11m/s = 1.1 seconds

    dx = vxt
    12.22m = vx(1.1seconds)
    vx = 11.1 m/s
    11.78m = vx(1.1seconds)
    vx = 10.7 m/s

    vx = v1cos35*
    11.1 m/s = v1cos35* = 14 m/s
    10.7 m/s = v1cos35* = 13 m/s

    range of initial speeds = 13-14 m/s
     
    Last edited: Oct 12, 2003
  2. jcsd
  3. Oct 14, 2003 #2
    I think the last one should read d =vyt + 1/2gt^2. It's OK since you used that in the next line:
    OK. I get v1 = 11.54 m/s.

    Why not plug in dx = 11.78m resp. dx = 12.22m into the same formula?
    This yields
    v1(min) = 11.44 m/s
    v1(max) = 11.63 m/s
     
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