Aball 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   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.

Ithink your answer is d