Question #18144
1 Answer
Explanation:
I'll assume the question says that the wall is
Well, since the ball covers a distance of
With that in mind, we'll use the equation
ul(Deltax = v_0cosalpha_0t
Since the time
color(red)(t = (Deltax)/(v_0cosalpha_0)
And for the
ul(v_y = v_0sinalpha_0 - g t
Solving for
color(red)(t = (v_0sinalpha_0 - v_y)/g
Setting two red equations equal to each other:
(Deltax)/(v_0cosalpha_0) = (v_0sinalpha_0 - v_y)/g
Now our only unknown is the initial speed
(2color(white)(l)"m")/(v_0cos(45^"o")) = (v_0sin(45^"o") - 0)/(9.81color(white)(l)"m/s"^2)
Solving for
color(green)(ul(v_0 = 6.26color(white)(l)"m/s"
Now that we know the initial speed, we can now find the maximum height using the equation
(v_y)^2 = (v_0sinalpha_0)^2 - 2g(Deltay)
Plugging in known values:
.
color(blue)(ulbar(|stackrel(" ")(" "Deltay = 1color(white)(l)"m"" ")|)
The height of the wall is thus