Question

Question #f7f80

Answer 1

(A) `29.4` `"m/s"`

(B) `44.1` `"m"`

Explanation:

We're asked to find

• (A) the initial speed of the ball

• (B) the maximum height of the ball

given its time of flight (`6.00` `"s"`).

`" "`

• (A) Initial speed

To find the necessary initial speed, we recognize that the particle will be at its maximum height at `t = 3.00` `"s"` (halfway split time between upward motion and downward motion), and at this point, the instantaneous `y`-velocity is `0`.

We can then use the equation

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

where

• `v_y` is the `y`-velocity at time `t` (`0`, maximum height)

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

• `g = 9.81` `"m/s"`

• `t` is the time (`3.00` `"s"`)

Plugging in known values:

`0 = v_(0y) - (9.81color(white)(l)"m/s"^2)(3.00color(white)(l)"s")`

`color(red)(ulbar(|stackrel(" ")(" "v_(0y) = 29.4color(white)(l)"m/s"" ")|)`

`" "`

• (B) Maximum height reached

To find this, we can use the kinematics equation

`Deltay = ((v_y + v_(0y))/2)t`

where

• `Deltay` is the change in height (what we're trying to find)

• `v_y` still equals `0` (maximum height)

• `v_(0y) = color(red)(29.4color(white)(l)"m/s"`

• `t = 3.00` `"s"`

Plugging in known values:

`Deltay = ((0+color(red)(29.4color(white)(l)"m/s"))/2)(3.00color(white)(l)"s")`

`color(blue)(ulbar(|stackrel(" ")(" "Deltay = 44.1color(white)(l)"m"" ")|)`