Question

Question #a06f3

Answer 1

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

`v_0` is the initial velocity of the ball
`h_"max"` is the maximum height it reaches

Rearrange to solve for `h_"max"`

`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 `"5 m"` 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 `h_"max"`, you can say that its velocity when it gets caught is equal to

`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 `h_"down" = "5 m"`, 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`

`{(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 `h_"down" = "5 m"` on its way down, the higher value for `t` will be correct.

Therefore, the trip took a total of `color(darkgreen)(ul(color(black)(t = "3.81 s"))`.

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.