Ask Question
16 October, 03:10

A ball is thrown with an angle of 25.0 to the horizon with a speed of 18.0 m/s. What are its horizontal and vertical components?

+5
Answers (1)
  1. 16 October, 03:18
    0
    Best Answer: vi=24 m/s

    θ=43°

    x=16m

    0) Use this equation because you have x component and not y.

    You need time or part c so you solve it like this

    x = (vi*cosθ) t

    16 = (24cos43°) t

    t=.911552s=.912s

    a) This is a trajectory and you need y so you use this equation.

    y = (tanθ) x - (gx^2 / (2 (vi*cosθ) ^2)

    y=tan43°*16m - ((-9.8*16m) ^2) / (2 (24*cos43° ...

    y=14.920m-39.901m

    y=-24.9808m = - 25.0m

    b) Break it down into it's components.

    vx=cosθ*vi

    vx=cos43°*24

    vx=17.5525m/s = 17.5m/s

    c) vy = sin 43°*24-.5 (-9.8) (.912s)

    vy=22.2m/s

    I'm not 100% positive on some of these but it should look something like the work above.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A ball is thrown with an angle of 25.0 to the horizon with a speed of 18.0 m/s. What are its horizontal and vertical components? ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers