A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground, another ball of mass m is thrown straight downward toward the first ball, also at speed v0. Assume that gravity acts vertically downward on each mass m with a magnitude mg. v0 D b How far above the ground do the balls collide (in terms of only D, v0, and g)?

Explanation:

Let the balls collide after time t .

distance covered by falling ball

s₁ = v₀ t + 1/2 g t²

distance covered by rising ball

s₂ = v₀ t - 1/2 g t²

Given ,

s₁ + s₂ = D

D = v₀ t + 1/2 g t² + v₀ t - 1/2 g t²

= 2v₀ t

t = D / 2v₀

s₂ = v₀ t - 1/2 g t²

= v₀ x D / 2v₀ - (1/2) x  g x D² / 4v₀²

= D / 2 - gD² / 8 v₀²

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.

c) power lifting, will be your answer.

,

deku ❤