Verified answer

first use trig

horizontal speed = 2000cos(83.9) = 212.53

vertical speed = 2000sin(83.9) = 1988.68

for flight time, s=ut + 0.5at^2 vertically

0=1988.68(t) + 0.5(-9.8)t^2

t(1988.68-4.9t)=0

so, 4.9t = 1988.68

t=405.9 s

distance horizontally = 405.9(212.53)=86265.2m

seems very far but 2000m/s is very fast

You're going to use the displacement equation

D = D0 + V0*t + 1/2*a*t^2

Since the beginning and ending are at ground level, both the initial displacement D0, and the displacement D are zero. "a" is -9.8 m/s^2 (it's negative because up is positive and g is downwards). The initial upward velocity is 2000 m/s times sin(83.9%) or 1989 m/s.

So, this equation becomes:

0 = 0 + 2000*sin(83.9)*t - 4.9*t^2

You can use the quadratic formula to solve this, or you can factor to get

0 = (2000*sin(83.9) - 4.9*t) * t

Because of this factoring, the equation is satisfied in two situations: t = 0 (when the shell is fired) or 4.9*t = 2000*sin(83.9). Or t = 2000*sin(83.9) / 4.9

I'll let you finish the arithmetic. Remember to convert from seconds to minutes. You're going to be around four hundred seconds, to give you something to compare against.

Now, for the horizontal range, you're going to start with the same equation:

D = D0 + V0*t + 1/2*a*t^2

However, in a projectile problem, there is no horizontal acceleration. The initial horizontal velocity is 2000*cos(83.9).

D = 0 + 2000*cos(83.9) * t + 0

You have the time from the previous problem. I'll let you finish up the arithmetic, but your answer should be somewhere around 80 or so.\

That's some freakin' cannon. It's throwing this shell some 200 km (120 miles) into the air.