Question

Question #e1645

Answer 1

`"height" = 24.5` `"m"`

`"velocity" = -29.4` `"m/s"`

Explanation:

We're asked to find

• the height of the tower

• the velocity at which the ball strikes the ground

from the given information.

`" "`

Height of Tower

This will be the initial position `y_0` of the ball, and we're going to use the kinematics equation

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

where

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

• `y_0` is the initial position (what we're trying to find)

• `v_(0y)` is the initial velocity of the ball (given as `19.6` `"m/s"`)

• `t` is the time of the motion (given as `5` `"s"`)

• `g = 9.8` `"m/s"^2`

Plugging in known values:

`0 = y_0 + (19.6color(white)(l)"m/s")(5color(white)(l)"s") - 1/2(9.8color(white)(l)"m/s"^2)(5color(white)(l)"s")^2`

`color(red)(ulbar(|stackrel(" ")(" "y_0 = 24.5color(white)(l)"m"" ")|)`

The `color(red)("height"` of the tower is thus `color(red)(24.5color(white)(l)"meters"`.

`" "`

Final Velocity

To find the final velcoity, 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 (`19.6` `"m/s"`)

• `g = 9.81` `"m/s"^2`

• `t` is the time (`5` `"s"`)

Plugging these in:

`v_y = 19.6color(white)(l)"m/s" - (9.8color(white)(l)"m/s"^2)(5color(white)(l)"s")`

`color(blue)(ulbar(|stackrel(" ")(" "v_y = -29.4color(white)(l)"m/s"" ")|)`

The final `color(blue)("velocity"` of the ball is therefore `color(blue)(29.4color(white)(l)"meters per second"` in the `color(blue)("downward direction"`.