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?

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.