A projectile is launched at an angle of 30° from the horizontal and lands 69 s later at the same height from which it was launched. (a) What is the initial speed of the projectile (in m/s)?

679.2 m/s

Explanation:

The formula for the time of flight of a projectile is given as,

T = 2usinФ/g Equation 1

Where T = Time of flight, u = initial velocity, Ф = angle of projectile to the horizontal, g = acceleration due to gravity.

make u the subject of the equation

u = Tg/2sinФ Equation 2

Given: T = 69 s, Ф = 30°

Constant: g = 9.8 m/s²

Substitute into equation 2

u = (69×9.8)/(2×sin30)

u = 679.2 m/s.

Hence the initial speed of the projectile is 679.2 m/s

hypothesising or postulating

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.