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  1. B

    Projectile motion without time and initial velocity

    Ok, I believe i found my source of error. When solving for initial velocity equation (7) should turn out as following: 7) Vi = sqrt( ((g)(xf^2)) / ((2)(cos^2(angle))((-yi) - (xf)(tan(angle)))) I left out the cos^2(angle). I redid the calculations and yielded the answer in the back of the...
  2. B

    Projectile motion without time and initial velocity

    Notice that my g = -9.8 m/s^2 . So the negative sign makes up for the presence of a positive g rather than a negative. I will redo my calculations now and get back to you.
  3. B

    Projectile motion without time and initial velocity

    So the given data is as follows: angle = 42 degrees yf = 0m yi = 2.1m xi = 0m xf = 17m I used the given data to produce the following six equations where "t" is time and "V" is velocity and "g" is the acceleration due to gravity -9.8m/s^2. 1) Vfx = (Vi)(cos(angle)) 2) xf =...
  4. B

    Projectile motion without time and initial velocity

    Homework Statement A shotputter projects the shot at 42 degrees to the horizontal from a height of 2.1 m. It lands 17 m away horizontally. Next, he gives it the same initial speed but changes the angle to 40 degrees. What effect does this have on the horizontal range? Homework...
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