webwork 10.9 question 6

I had a question about chapter 10.9 number 6. Ive tried multiple tries
myself and I have even gone to tutoring centers and they say the work is
right but webwork marks it as wrong. I am hoping you could guide me on how
to solve this problem.

The acceleration due to gravity is a=<0,-9.8>.  The Sin(30˚)=1/2, Cos(30˚)=√3/2, so the initial velocity vector is v_0=250<√3/2,1/2>=<125√3,125>. Since a(t)=v'(t), and a(t)=a is constant, it follows that

v(t)=a*t+v_0=<125√3,125-9.8*t>.  

Since v(t)=r'(t), it follows that

r(t)=a*t^2/2+v_0*t+r_0.  

But since the projectile is fired from the ground, and it's a reasonable thing to make our coordinate system so that the origin is at (0,0), we can assume that r_0=0, hence

  r(t)=a*t^2/2+v_0*t=<125√3*t,125*t-9.8*t^2/2>.

Writing r(t)=<x(t),y(t)>, the projectile will be at ground level when y(t)=0, i.e. when 

0=125*t-9.8*t^2/2=t(125-9.8*t/2).

This equation has two solutions: t=0, before the projectile is fired, and t=250/9.8=25.51 when the particle hits the ground again.  At that time x(t)=125√3*25.51=5523.1 meters, so the range is 5523.1. At the time the particle hits the ground, the speed will be the magnitude of the vector 

||v(25.51)||=||<125√3,125-9.8*t>||=||<125√3,125-9.8*25.51>||=||<125√3,-125>||=250.

The maximum height will be attained with the vertical component of the velocity is zero, i.e. when 

0=125-9.8*t  or t=125/9.8=12.76.

At this time the height is y(12.76)=125*12.76-9.8*(12.76^2)/2=796.6 meters.