From the top of a building 85.0 meters tall a ball is dropped. at the same time another ball is thrown upward from the ground with a speed of 46.0 m/s. a. how long after the balls are released will they hit? b. how high above the ground will these two balls hit?

a) Both balls are affected by gravity in the same way. The time both balls hit each other is equal to the time the ball thrown up reaches the top of the roof, if there was no gravity. (both balls would be at the same location)

y = v*t = 85m = 46m/s * t

t = 1.85s

Check:

The equation for position is given by:

y(t) = -0.5gt² + v₀t + x₀

The position of the ball on the roof top at time t = 1.85s:

y₁(t) = -4.9t² + 85.0 = 68.3m

The position of the ball thrown up at time t = 1.85s:

y₂(t) = -4.9t² + 46 * 1.85 = 68.3m

b) the two balls hit each other at a hight of 68.3m

im guessing the west of station b

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.