Acannon fires a 0.652 kg shell with initial
velocity vi = 12 m/s in the direction = 61 ◦
above the horizontal.
the shell’s trajectory curves downward be-
cause of gravity, so at the time t = 0.473 s
the shell is below the straight line by some
vertical distance h.
find this distance h in the absence of
air resistance. the acceleration of gravity is
9.8 m/s2 .
answer in units of m

height= 6.429m

Explanation:

h is equall to Max height traveled by trajectory body.

Max height= (U²Sin²tita)/2g

Max height = (12²(sin61)²)/(2*9.8)

Max height =( 144*0.875)/19.6

Max height= 6.429 m

less than one you can't create energy

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.