# Question #a06f3

##### 1 Answer

Here's what I got.

#### Explanation:

The first thing to do here is to calculate the **maximum height** that the stone reaches.

At the top of its trajectory, i.e. at *maximum height*, its velocity will be equal to **zero**, which means that you can write

#0 = v_0^2 - 2 * g * h_"max"#

Here

Rearrange to solve for

#v_0^2 = 2 * g * h_"max" implies h_"max" = v_0^2/(2 * g)#

Plug in your values to find

#h_"max" = (20^2 "m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-2)))))/(2 * 9.81 color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s"^(-2))))) = "20.4 m"#

Now, you know that the stone was caught **on its way down** at a height of **above** the point from which it was thrown.

This essentially means that the stone traveled a total of

#h_"down" = "20.4 m" - "5 m" = "15.4 m"#

on its way down from **its maximum height**. Since its velocity was equal to zero at

#v_"caught"^2 = 0^2 + 2 * g * h_"down"#

Plug in your values to find

#v_"caught" = sqrt(2 * "9.81 m s"^(-2) * "15.4 m") = color(darkgreen)(ul(color(black)("17.4 m s"^(-1))#

To find the time of the trip, we can use the fact that the stone is passing through the point at which it gets caught, i.e. through **twice**, once on its way up and once on its way down.

You can thus say that you have

#h_"down" = v_0 * t^2 - 1/2 * g * t^2#

Plug in your values and rearrange the equation as

#4.905 * t^2 - 20 * t^2 + 5 = 0#

This quadratic equation will produce two values for

#{(t_1 = "3.81 s"" "color(darkgreen)(sqrt())), (t_2 = color(red)(cancel(color(black)("0.268 s")))) :}#

Since we're looking for the time needed for the stone to pass through **down**, the *higher value* for

Therefore, the trip took a total of

I'll leave both answers rounded to three **sig figs**, but keep in mind that you only have one sig fig for the initial velocity and height at which the stone is caught.