How far will the bullet travel before falling back to earth? First, the question:

A particular high performance rifle cartridge can be fired with a muzzle velocity of 1200 meters per second. The rifle is pointed straight up. Assuming there is no air resistance, how high does the rifle bullet travel before it begins to fall back to earth?

Now, I have a feeling this will require an integral, as the velocity of the bullet will change slowly due to the force of gravity acting upon it in a negative direction. I'm not sure how to set this integral up, however. I thought perhaps it was this:
$$\int_0^\infty (m_{earth} \times v) - 9.81dv$$
But after looking at that, it just makes no sense at all. Can anyone guide me in the right direction?
 A: This problem does not in fact require an integral. Any object in freefall (below about 100 km) has a constant acceleration due to the constant force of gravity. If you actually performed this experiment (don't) then air resistance would play a significant role. However, ignoring air resistance, the kinematic equation for relating velocity to position with constant acceleration is $v_{f,y}^2 = v_{i,y}^2+2a(y_f-y_i)$. The highest point is a turning point, i.e. the bullet turns around. This means that the bullet's instantaneous velocity $v_{f, y}$ at the highest point is $0~\rm m/s$. The bullet's initial velocity $v_{i, y}$ is known to be $1200~\rm m/s$. The acceleration due to gravity is $-g=-9.8~\rm m/s^2$. The initial position $y_i$ is known to be 0 m, and the final position $y_f$ is what we are looking for. The new equation is $0~{\rm m^2/s^2} = 1440000~{\rm m^2/s^2}-19.6~{\rm m/s^2}*y_f$. Hence $y_f$ equals $(1440000/19.6)~{\rm m} = 73500~\rm m$. This means our initial assumption about the constant force of gravity is inaccurate by less than 0.01 %, effectively 0. Anyway, try to learn the 3 kinematic equations for constant acceleration, they are very useful. By the way, it may interest you to know that these equations are actually derived from integrating the velocity function. I hope this helped answer your question, and have a nice day.
