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Astronaut Ballplayer

  1. Apr 8, 2005 #1
    Q. An astronaut in his space suit can throw a ball a maximum distance dmax = 8 m on the surface of the earth.

    a) For a given speed of the ball, what angle to the horizontal q (in degrees) will yield the greatest range?
    q = ° 45 *

    b) If the ball is thrown at this same angle q, what speed will produce this greatest range (8 m) ?
    v = m/s *
    sqrt (78.48) OK

    c) How far can he throw the ball on a planet where g1 = 19 m/s2?
    xp = m *
    9.81*8/19 OK

    d) What height will the ball reach on this "maximum range" trajectory? (on the planet where g1 = 19 m/s2)
    hmax = m
    3.91 NO

    Need help with part d,

    I think we need to use the formula , X = V sqrt (2H/g), if yes, what to plug in,

    Pl. Help
  2. jcsd
  3. Apr 8, 2005 #2
    for d
    Vertical ONLY
    what is your initial velocity COMPONENT that points upward? that's v1. What is the velocity of the ball at it's maximum height? what is the acceleration?
    use this formula
    [tex] v_{2}^2 = v_{1}^2 + 2ad [/tex]
    and dont forget about a sign convention, take one direction (up or down) to be positive and the other to be negative. You cannot take the square root of a negative number - you have toi get a REAL number
  4. Apr 9, 2005 #3
    At the maximum height, v2 = 0

    so, 2ad = - v1^2

    d = -v1^2/2(-9.81)

    v1= v1sin(45)

    I get 1.027m
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