A cannon launches the cannon ball to the maximum horizontal distance of 3279 ft. Assuming no air resistance, and assuming the cannon ball is launched at ground level, what is the minimum initial speed of the cannon ball (just as it leaves the cannon) that is needed for it to reach this distance, in m/s

The minimum initial speed of the cannon ball is approximately 99.018 m/s

Explanation:

The parameters of the cannon are;

The maximum horizontal distance reached, R = 3,279 ft. ≈ 999.4392 m

The maximum horizontal range, , is given by the following formula;

Where;

u = The initial speed

g = The acceleration due to gravity ≈ 9.81 m/s²

= 999.4392 m

We get;

u = √(g × )

∴ u = √(9.81 m/s² × 999.4392 m) ≈ 99.018 m/s

The minimum initial speed of the cannon ball (just as it leaves the cannon) that is needed for it to reach this distance, in m/s, u ≈ 99.018 m/s.

because growing process had stoped after 4-5 year

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

where is unknown. because the position is parabolic, there is symmetry about the midpoint at , and we know that

so that the position function is

which means the velocity is

then the total distance traveled is

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at and ends at , so its displacement is 0.