How far will a pellet travel

[Harry's Lad]

I should have studied harder in school....I have no idea what any of you are talking about

Here, does this help?

Maximum range achieved at an angle of 45 degrees . . .

That would be true in a vacuum (no drag – infinite BC) . . . in the atmosphere though (especially with low BC, high-drag airgun pellets), the optimum angle is a lot less.

out of interest then, if you were to shoot vertically up, the pellet would eventually stop and fall down, so what would be its terminal velocity?!

The free-fall terminal velocity is about 880 * sqrt(BC) Ft/s.
0.177 JSB Exact Vt = ~ 880 * sqrt(0.021) = ~ 127 Ft/s.
With the more manly calibre (0.22 JSB Exact say) Vt = 880 * sqrt(0.032) = ~ 157 Ft/s.

i mean i assume it would be the same as if you just dropped it fform a great height!?

Correct since it is indeed being dropped from a, er, great height.

always pondered that one, but i dont get out much!

Yeah, me too.

Somebody will correct me, but as to pellet size .22 will travel further than .177 as if you throw a squash ball and a ping pong ball the squash ball will go further.... does that make sense? lol

Maximum range is a function of initial velocity and Ballistic Coefficient alone (and the ability to calculate the optimum angle of course). Calibre, mass and moon-phase are irrelevant.
The quash ball has a much greater BC (than the ping pong ball) because of its vastly superior Sectional Density (since BC = Sectional Density/Form Factor and the Form Factor is the same for both balls) so would travel further at any non-zero angle and have a greater Terminal Velocity.

If you use the idea that the pellet is fired directly upwards then it will accelerate upwards slowing down until it stops, it will then fall. As it falls it will gain speed until it reaches terminal velocity. That is 9.81m/s/s. Force is mass multiplied by acceleration so the heavier the pellet the more force it has at terminal velocity (which would be the same).

At the top of the trajectory the projectile’s velocity is zero (relative to the ground). As it descends, it’s downward velocity increases driven by gravitational acceleration (~32.17 Ft/s^2 as you say). However, as the velocity increases it generates drag – proportional to velocity squared – the vector of which is acting against the gravitational acceleration. As the projectile’s downward velocity increases, the drag increases at an exponential rate until a limiting velocity is reached such that the drag force exactly equals the accelerating force (gravity). This equilibrium is reached at approx. 880 * sqrt(BC) [Ft/s] as above.

Why do we use big bullets for deer?

Big bullet = big mass = big kinetic energy . . . and big mass = big sectional density = big BC means that a bigger proportion of the initial KE and momentum are available at the target.

In case you were wondering . . .

Maximum_Range = 11000 * BC * Loge(MV / (880 * Sqrt(BC))) [Yard]

Optimum_Angle = 45 * (1 - Exp(-((11000000 * BC / MV^2)) ^ 0.333)) [Degree]

. . . believed to be the original work of Steve Woodward (Steve_in_NC)

HTH
Dave