A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground

t = 2 s

Explanation:

In order to find the time taken by the stone to fall from the top of the building to the ground we can use 2nd equation of motion. 2nd equation of motion is as follows:

s = Vit + (0.5)gt²

where,

t = time = ?

Vi = Initial Velocity = 20 m/s

s = height of building = 60 m

g = 9.8 m/s²

Therefore,

60 m = (20 m/s)t + (0.5)(9.8 m/s²)t²

4.9t² + 20t - 60 = 0

solving this quadratic equation we get:

t = -6.1 s   (OR)   t = 2 s

Since, the time cannot be negative in magnitude.

Therefore,

t = 2 s

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.