Question

Question #c287b

Answer 1

(A) `"max height" = 87.9` `"m"`

(B) `"final velocity" = -41.5` `"m/s"`

(C) `"time" = 0.767` `"s"`

Explanation:

We're asked to find

• (A) the maximum height the stone reaches

• (B) the velocity at which the stone strikes the ground

• (C) the time taken to reach the maximum height

`" "`

• (A) Maximum Height

To find the maximum height obtained by the stone (and all other information we need), we'll first find the initial velocity of the stone, using the equation

`ul(y = y_0 + v_(0y)t - 1/2g t^2`

where

• `y` is the final position (`0`, ground level)

• `y_0` is the initial position (`85` `"m"`)

• `v_(0y)` is the initial velocity (what we're trying to find)

• `t` is the time (`5` `"s"`)

• `g = 9.81` `"m/s"^2`

Plugging in known values:

`0 = 85` `"m"` `+ v_(0y)(5color(white)(l)"s") - 1/2(9.81color(white)(l)"m/s"^2)(5color(white)(l)"s")^2`

`1/2(9.81color(white)(l)"m/s"^2)(5color(white)(l)"s")^2 = 85color(white)(l)"m" + v_(0y)(5color(white)(l)"s")`

`color(red)(v_(0y)) = (1/2(9.81color(white)(l)"m/s"^2)(5color(white)(l)"s")^2 - 85color(white)(l)"m")/(5color(white)(l)"s") = color(red)(ul(7.53color(white)(l)"m/s"`

At the stone's maximum height, its instantaneous `y`-velocity is zero, so we can use the equation

`ul((v_y)^2 = (v_(0y))^2 - 2g(y - y_0)`

The variables here we know except for the final height `y`, which will be

`0 = (color(red)(7.53color(white)(l)"m/s"))^2 - 2(9.81color(white)(l)"m/s"^2)(y - 85color(white)(l)"m")`

`color(blue)(ulbar(|stackrel(" ")(" "y = 87.9color(white)(l)"m"" ")|)`

`" "`

(B) Final Velocity

To find the final velocity, we can use the equation

`ul(v_y = v_(0y) - g t`

where

• `v_y` is the final velocity (what we're trying to find)

• `v_(0y)` is the initial velocity (found to be `color(red)(7.53color(white)(l)"m/s"`)

• `g = 9.81` `"m/s"^2`

• `t = 5` `"s"`

Plugging these in:

`color(blue)(v_y) = color(red)(7.53color(white)(l)"m/s") - (9.81color(white)(l)"m/s"^2)(5color(white)(l)"s") = color(blue)(ulbar(|stackrel(" ")(" "-41.5color(white)(l)"m/s"" ")|)`

`" "`

• (C) Time at Maximum Height

To find the time when the stone reaches its maximum height, we can again use the equation

`ul(v_y = v_(0y) - g t`

where

• `v_y = 0` (at maximum height)

• `v_(0y) = color(red)(ul(7.53color(white)(l)"m/s"`

• `g = 9.81` `"m/s"`

• `t` is the time (what we're trying to find)

And so we have

`0 = color(red)(7.53color(white)(l)"m/s") - (9.81color(white)(l)"m/s"^2)t`

`color(blue)(ulbar(|stackrel(" ")(" "t = 0.767color(white)(l)"s"" ")|)`