Horizontal Distance of Cannon Fired at 30 Degrees: 85,000m

[ucdawg12]

Here is the question: A large cannon is fired over level ground at an angle of 30 degrees above the horizontal. The muzzle velocity is 980m/s. What horizontal distance (in meters) will the projectile travel before striking the ground?

I know the answer, it is 85,000 meters, however I am having trouble arriving at that answer myself. Here is what I am doing. First I find how long it is in the air, I find the 980sin(30) to find the vertical velocity of 490. then I use the v=v(initial) + at, 0 = 490 +(-9.8t), t=50. Then to find the horizontal distance I do 980cos(30) = 850. Then v=x/t, vt=x, 850(50)=43000. But that isn't the right answer. Does anyone know what I did wrong? Thanks

 

[vincentchan]

you are basically right, BUT, who told you the final velocity is zero? could it be zero in your case? if not, what is it?

 

[ucdawg12]

oh god I just realized what I did, right, when the velocity was 0 the t was 50seconds but that only meant that it took 50 seconds to reach its peak, it would take another 50 seconds for it to come back down so it is 850(100)=x=85000m.

thanks :)