Question

Question #248e7

Answer 1

`v_"final" = 20.8` `"m/s"`

Explanation:

We're asked to find the speed of the ball as it reaches the ground, given its initial height and initial velocity.

We're given that its initial height is `12.2` `"m"`, and that it was kicked solely horizontally, so there is no initial `y`-velocity.

We know the horizontal speed remains ideally unchanged throughout the motion, so we need to find the final `y`-velocity of the ball when it hits the ground.

To do so, we can use the equation

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

where

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

• `v_(0y)` is the initial `y`-velocity (`0` because the initial velocity was horizontal)

• `g = 9.81` `"m/s"`

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

• `y_0` is the initial position (`12.2` `"m"`)

Plugging in known values:

`(v_y)^2 = 0 - 2(9.81color(white)(l)"m/s"^2)(0-12.2color(white)(l)"m")`

`v_y = +-sqrt(239color(white)(l)"m"^2"/s"^2) = color(red)(15.5color(white)(l)"m/s"`

Since we're dealing with speed and not velocity, I neglected the sign, because speed is not a vector quantity.

Now, we use the Pythagoream theorem to find the final speed given the components:

• `v_x = 13.9` `"m/s"` (`x`-motion doesn't change in idealized projectile motion)

• `v_y = color(red)(15.5color(white)(l)"m/s"`

And so we have

`color(blue)(v) = sqrt((v_x)^2 + (v_y)^2) = sqrt((13.9color(white)(l)"m/s")^2 + (color(red)(15.5color(white)(l)"m/s"))^2) = color(blue)(ulbar(|stackrel(" ")(" "20.8color(white)(l)"m/s"" ")|)`

which agrees with the answer you gave us.