Question #8658b

The stone is being thrown down from a height of #"4.4 m"# 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 #d# to be the depth of the well, #v_0# to be the initial velocity of the stone, and #v# to be the 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#

#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, #g#, is accelerating the velocity of the stone by #"9.81 m s"^(-1)# 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 #t_0=0# to be the starting time, i.e. the time at which the stone was thrown, you can say that

#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 #t# -- I won't add the units to keep things simple

#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 #t = "0.70 s"# as the answer.