Question

Question #ae3f1

Answer 1

`v_0 = 28.0` `"m/s"`

Explanation:

We're asked to find the speed at which the ball was hit, given its horizontal range and hit angle.

To do this, we can use the equations

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

`ul(Deltax = v_0cosalpha_0t`

where

• `Deltay` is the change in height (`0` since the ball landed at the same height it was hit)

• `Deltax` is the change in horizontal position, given as `103` `"m"`

• `v_0` is the initial speed (what we're trying to find

• `alpha_0` is the hit angle, given as `50^"o"`

Plugging in known values to both equations:

`0 = v_0sin(50^"o")t - 1/2(9.81color(white)(l)"m/s"^2)t^2`

`103color(white)(l)"m" = v_0cos(50^"o")t`

We can realize that the time `t` when the ball is at `y`-position `0` and `x`-position `103` `"m"` are equal, so we can solve the second equation for `t` and plugbthat in for `t` in the first equation (and eliminate units to make it simpler):

`t = (103)/(v_0(0.643)) = 160.2/(v_0)`

`0 = v_0(160.2/(v_0)) - 1/2(9.81)(160.2/(v_0))^2`

Now we solve for `v_0`:

`4.905(160.2/(v_0))^2 = 160.2`

`125944/((v_0)^2) = 160.2`

`v_0 = sqrt((125944)/(160.2)) = color(red)(ulbar(|stackrel(" ")(" "28.0color(white)(l)"m/s"" ")|)`