Answer:45.24 m/s
Explanation:
Given
Height of Building h=24 m
Range of ball R=100 m
Considering Vertical motion of ball
using [tex]y=u_yt+\frac{a_yt^2}{2} [/tex]
initial vertical velocity is zero therefore [tex]u_y=0[/tex]
[tex]24=0+\frac{9.8\times t^2}{2}[/tex]
[tex]t=\sqrt{\frac{48}{9.8}}[/tex]
[tex]t=2.21 s[/tex]
Now considering Horizontal Motion
[tex]R=u_xt+\frac{a_xt^2}{2}[/tex]
[tex]100=u_x\times 2.21+0[/tex] , as there is no horizontal acceleration
[tex]u_x=45.24 m/s[/tex]
