# Question #8658b

##### 1 Answer

#### Answer:

#### Explanation:

The stone is being **thrown down** from a height of *above the bottom* of the well, so right from the start, you know that gravity is going to **accelerate** this stone, not decelerate it.

Let's try to solve this using an *intuitive* approach first. If you take **final velocity** of the stone, i.e. the velocity with which it hits the bottom, you can say that

#v^2 = v_0^2 + 2 * g * d#

Rearrange to solve for

#v = sqrt(v_0^2 + 2 * g * d)#

Plug in your values to find

#v = sqrt(2.8^2 "m"^2 "s"^(-2) + 2 * "9.81 m s"^(-2) * "4.4 m")#

#v = "9.70 m s"^(-1)#

Now, you know that the gravitation acceleration, **accelerating** the velocity of the stone by **with every passing second**.

This means that all you have to do to find the time travelled by the stone is use the *definition* of acceleration, which is **change in velocity** divided by **change in time**.

In this case, you'd have

#g = (Deltav)/(Deltat) = (v - v_0)/(t - t_0)#

If you take

#g = (v - v_0)/t implies t = (v - v_0)/g#

Plug in your values to find

#t = (9.70 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))) - 2.8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(9.81 color(red)(cancel(color(black)("m"))) "s"^color(red)(cancel(color(black)(-2)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("0.70 s")color(white)(a/a)|)))#

The answer is rounded to two **sig figs**.

Let's double-check the result by using a *single equation*

#d = v_0 * t + 1/2 * g * t^2#

Plug in your values to find

#"4.4 m" = "2.8 m s"^(-1) * t + 1/2 * "9.81 m s"^(-2) * t^2#

Rearrange to solve for

#4.905 * t^2 + 2.8 * t - 4.4 = 0#

This quadratic equation will produce *two solutions*, one positive and one negative.

#{(t = "0.70 s"" "color(green)(sqrt())), (t_2 = color(red)(cancel(color(black)(-"1.3 s")))) :}#

The negative one has no physical meaning in this context, which means that you're once again left with