Physics, 12.09.2019 04:30 marleacorey

While standing on a bridge 15.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.20 m, you throw a second stone straight down. what initial velocity must you give to the second stone if they are both to reach the ground at the same instant?

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Answer from: aliami0306oyaj0n

11.35 m/s

Explanation:

Let the first ball takes time t to reach the ground.

Use second equation of motion

$s=ut + 0.5at^{2}$

Here, s = 15 m, a = 9.8 m/s^2, u = 0 m/s

By substituting the values

15 = 0 + 0.5 x 9.8 x t^2

t = 1.75 second

Now time taken by the first ball to reach 3.20 m is t'

Use second equation of motion

$s=ut + 0.5at^{2}$

Here, s = 3.2 m, a = 9.8 m/s^2, u = 0 m/s

By substituting the values

3.2 = 0 + 0.5 x 9.8 x t'^2

t' = 0.81 s

Thus, the time for the second ball to reach the ground is t - t' = 1.75 - 0.81 = 0.94 s

For second ball:

Use second equation of motion

$s=ut + 0.5at^{2}$

Here, s = 15 m, a = 9.8 m/s^2, u = ?, t = 0.94 s

By substituting the values

15 = u x 0.94 + 0.5 x 9.8 x 0.94 x 0.94

u = 11.35 m/s

Thus, the second ball has the initial velocity of 11.35 m/s.

Answer from: Quest

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

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

not too sure thinking all accelerations are pointed down

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While standing on a bridge 15.0 m above the ground, you drop a stone from rest. when the stone has f...

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