Question #248e7

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.