Question

Question #2027f

Answer 1

(i) `7.96` `"m"`

(ii) `0.24` `"s"`

(iii) `0.375` `"s"` and `2.17` `"s"`

Explanation:

We're asked to find

• (i) the maximum height obtained by the mass

• (ii) the time it takes the mass to travel `3.0` `"m"` horizontally

• (iii) the time(s) when the height of the mass is `4.0` `"m"`

(i)

Finding the maximum height:

When the particle is at its maximum height, the instantaneous `y`-velocity is `0`, and we can use the kinematics equation

`ul((v_y)^2 = (v_0sinalpha_0)^2 - 2g(Deltay)`

where

• `v_y` is the `y`-velocity at height `Deltay` (which is `0`)

• `v_0` is the initial speed (given as `25` `"m/s"`)

• `alpha_0` is the launch angle (given as `30^"o"`)

• `g = 9.81` `"m/s"^2`

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

Plugging in known values:

`0 = [(25color(white)(l)"m/s")sin(30^"o")]^2 - 2(9.81color(white)(l)"m/s"^2)(Deltay)`

`color(red)(ulbar(|stackrel(" ")(" "Deltay = 7.96color(white)(l)"m"" ")|)`

The maximum height is thus `color(red)(7.96color(white)(l)"meters"`.

`" "`

(ii)

The time it takes to travel `3.0` `"m"` horizontally:

To do this, we can use the equation

`ul(Deltax = v_0cosalpha_0t`

where

• `Deltax` is the change in horizontal position (`3.0` `"m"`)

• `v_0 = 25` `"m/s"`

• `alpha_0 = 30^"o"`

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

Plugging in known values:

`3.0color(white)(l)"m" = (25color(white)(l)"m/s")cos(30^"o")t`

`color(red)(ulbar(|stackrel(" ")(" "t = 0.24color(white)(l)"s"" ")|)`

It thus takes `color(red)(0.24color(white)(l)"seconds"` to travel `3.0` `"m"` horizontally.

`" "`

(iii)

The time(s) when the height is `4.0` `"m"`:

We can now use the equation

`ul(Deltay = v_0sinalpha_0t - 1/2g t^2`

where

• `Deltay` is the change in vertical position (`4.0` `"m"`)

• `v_0 = 25` `"m/s"`

• `alpha_0 = 30^"o"`

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

• `g = 9.81` `"m/s"^2`

Plugging in known values:

`4.0color(white)(l)"m" = (25color(white)(l)"m/s")sin(30^"o")t - 1/2(9.81color(white)(l)"m/s"^2)t^2`

Using the quadratic formula yields

`color(red)(ulbar(|stackrel(" ")(" "t = 0.375color(white)(l)"s"color(white)(l)"and"color(white)(l)t = 2.17color(white)(l)"s"" ")|)`

Therefore, the mass has a height of `4.0` `"m"` when the time is `color(red)(0.375color(white)(l)"seconds"` and `color(red)(2.17color(white)(l)"seconds"`.