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Solving for time

  1. Feb 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball is launched from a height of 1.9m with a speed of 40 m/s at an angle of 10 degrees above the horizontal. Find the total time in the air.

    2. Relevant equations
    ## y_f = y_0 + v_0y t + \frac{1}{2} t^2 ##

    3. The attempt at a solution
    Taking y0 = 1.9 m and v0 = 40sin(10), then using the quadratic formula, I got the answers t = 1.65 or t = -0.23. Clearly the former is what we need. However, my solutions guide got t = 3.3 instead (that is, double my answer). The solution guide used the equation ## t = \frac{-v_0 sin(θ) - \sqrt{v_0^2 sin^2(θ) + 2gy_0}}{-\frac{1}{2}g} ## and got 3.3 seconds. Why is this the case?
     
  2. jcsd
  3. Feb 19, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    They failed to multiply the coefficient of the squared term by 2 in the denominator. For the quadratic ax2 + bx + c = 0, the quadratic formula puts "2a" in the denominator.
     
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