Question #248e7
1 Answer
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
We know the horizontal speed remains ideally unchanged throughout the motion, so we need to find the final
To do so, we can use the equation
ul((v_y)^2 = (v_(0y))^2 - 2g(y - y_0)
where
-
v_y is the finaly -velocity (what we're trying to find) -
v_(0y) is the initialy -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.