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A quarterback throws the football to a stationary receiver who is 18.3 m down the field. Football is thrown at an angle of 44.9 degrees to the ground. Acceleration is 9.81 m/s2. What initial spped must the quarterback throw ball to reach receiver and what is the highest point during it's flight.

  • physics -

    The horizontal velocity is Vi*cos44.9 and the vertical velocity is Vi*sin44.9

    horizontal
    18.3=Vi*cos44.9 * time find time in terms of Vi.

    Vertical
    hf=hi+Vi*sin44.9*t -1/2 g t^2
    now hf=hi=0
    so
    0=Vi*sin44.9 -1/2 g t and put the expression you got for t, then solve for Vi. You will encounter the need for one trig identity.

  • physics -

    Split the problem in two, rising and falling. It does half the time and distance rising. (parabolic path)
    Initial speed = S
    U = constant speed in x direction = S cos 44.9 = .708 s
    Vo = initial speed in y direction = S sin 44.9 = .706 s
    vertical speed at max height = 0
    so
    0 = Vo - 9.81 t
    where t is time rising (half of total time in the air)
    t = .706 s/9.81 = .0720 s
    In half the time in the air (t) it goes half the distance down field , 9.15 m
    U t = 9.15
    t = 9.15 /(.708 s)
    so
    .072 s = 9.15 / (.708 s)
    .051 s^2 = 9.15
    s^2 = 180
    s = 13.4 m/s
    now how high?
    t = 9.15/(.708 s) = 9.15/(.708*13.4) = .964 seconds rising
    h = Vo t - 4.9 t^2
    h = .706*13.4 *.964 - 4.9 (.964^2)
    = 4.56 meters (about 14 feet)

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