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Projectile Motion with Basketball

  1. Jun 1, 2011 #1
    A person walking with a speed of 1.45 m/sec releases a ball from a height of h = 1.3 m above the ground. Use the point on the ground, directly below where the ball is initially released at the origin of your coordinate system.

    Picture: https://wug-s.physics.uiuc.edu/cgi/...k/Ch-03-04/wt-bouncing-ball/bouncing-ball.jpg


    (a) What is the ball's position at 0.25 seconds after it is released in the x and y direction?
    x =
    y =

    (b) What is the ball's position at 0.5 seconds after it is released in the x and y direction?
    x =
    y =

    (c) What is the ball's total velocity, speed and direction of motion at 0.5 seconds after it is released?
    vx =
    vy =
    v =
    Θ =

    For a, I tried using vf= vi + at. I used gravity (9.81) for the acceleration. However, its not right. For part b, I understand that you would use the same equation as part A, but you would just switch the 0.25 seconds to 0.5 seconds. For part c, I understand you use sin and cosine in order to solve for vx and vy.
     
  2. jcsd
  3. Jun 1, 2011 #2
    a) Finding the horizontal distance after 0.25 s is fairly straightforward:

    Given: vx = 1.45 m/s

    dx = vxt
    = (1.45 m/s)(0.25 s) = 0.36 m

    Finding the vertical height at 0.25 s:

    Given: v1y = 0

    dy = 0 + 0.5a(t)^2
    = 0.5(-9.8)(0.25)^2
    = -0.31 m

    1.3 m - 0.31 m = 0.99 m

    b) ... same process

    c) dy = 0 + 0.5at^2 = 0.5(-9.8)(0.5)^2
    = -1.23 m

    v2y2 - v1y2 = 2aydy
    v2y2 = 2(-9.8)(-1.23) + (0)
    v2y = 4.9 m/s
    v2x = 1.45 m/s (horizontal speed is constant throught)

    Then do a Pythagorean triangle:

    v^2 = (4.9)^2 + (1.45)^2
    v = 4.8 m/s

    tanx = 4.9/1.45
    x = 73.5 degrees below the horizontal
     
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