Physics, 02.10.2021 15:00, ciarrap552

A projectile is launched vertically into the air with a velocity of 19.0 m/s. Calculate the time it takes for the projectile to return to its starting height.
Pleas hur

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answered: jaemitchell23

Explanation:

vi = 19m/s

vf = 0

a = - 9.81

t = ?

Notice the minus sign on the acceleration. Either Vi or the acceleration has to be negative because they go in opposite directions.

equation

a = (vf - vi)/t

Solution

- 9.81 = (0 - 19)/t Multiply both sides by t

-9.81 * t = - 19.0 Divide both sides by 9.81

t = -19.0/9.81

t = 1.94 seconds

But that is the peak of the flight. It takes the same amount of time to descend and hit the ground.

Total time = 2 * 1.94

Total time = 3.88 seconds.

There's another way that you could do this.

Givens

vi = 19 m/s

vf = - 19 m/s

a = - 9.81

t = ?

a = (vf - vi)/t

-9.81 = (-19 - 19) / t

-9.81 * t = - 38

t = - 38/-9.81

t = - 3.87

What you are saying when you do the problem this way, is that what goes up must come down. It assumes that the landing speed is the same direction as the acceleration. It has the virtue of being shorter.

answered: Guest

answer: b.

explanation:

answered: Guest

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.

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A projectile is launched vertically into the air with a velocity of 19.0 m/s. Calculate the time i...