Question #18144

I'll assume the question says that the wall is #2# #"m"# horizontally from the point where the ball is thrown, and we're asked to find the height of the wall. I'll also assume you meant to say a distance of four meters from the throwing point is where it lands.

Well, since the ball covers a distance of #4# meters, and the wall is located at #2# #"m"# (i.e. in the center of its motion), then the ball just clears the wall at its maximum height (because in idealized projectile motion, the particle covers equal distance before and after it reaches its maximum height, which is directly in the middle of its motion).

With that in mind, we'll use the equation

#ul(Deltax = v_0cosalpha_0t#

Since the time #t# is the same whenever it has horizontal distance #2# #"m"# and vertical distance equal to its maximum height, we'll solve two equations for #t# (one for #x# and one for #y#), and set them equal to each other:

#color(red)(t = (Deltax)/(v_0cosalpha_0)#

And for the #y#-component, we'll use the equation

#ul(v_y = v_0sinalpha_0 - g t#

Solving for #t#:

#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 #v_0#, which we'll now find:

#(2color(white)(l)"m")/(v_0cos(45^"o")) = (v_0sin(45^"o") - 0)/(9.81color(white)(l)"m/s"^2)#

Solving for #v_0# yields

#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:

. #0 = [(color(green)(6.26color(white)(l)"m/s"))sin(45^"o")]^2 - 2(9.81color(white)(l)"m/s"^2)(Deltay)#

#color(blue)(ulbar(|stackrel(" ")(" "Deltay = 1color(white)(l)"m"" ")|)#

The height of the wall is thus #color(blue)(1color(white)(l)"meter"#.