# How far will the bullet travel before falling back to earth? [closed]

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?

• You're right: it doesn't. Try and look up 'SUVAT equations'.
– Gert
Oct 22 '15 at 17:22
• Why do you have 9.81 in the integral? Why are you integrating the velocity to $\infty$.? Oct 22 '15 at 17:28
• If I were you I'd try to solve that using conservation of energy: the loss in kinetic energy will be compensated by the increase in gravitational potential energy. So try to find the kinetic and potential energy at the different points. Oct 22 '15 at 17:59
• A shortcut you can use in this kind of problem is to work with escape velocity as a proxy for energy (although that might not be justified in a homework problem). Earth escape velocity at the surface is 11200 m/s. If you find the height at which the escape velocity equals 11200 m/s - 1200 m/s = 10000 m/s, that's the height the bullet will reach. Even if it's not an allowable solution, it's a way to check your work. Oct 22 '15 at 19:45
• Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. Oct 23 '15 at 8:55

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.