Answer:16.59 m
Explanation:
Given
initial horizontal speed of ball(u)=15 m/s
Height of building =6 m
Consider vertical motion first
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
here initial vertical velocity is zero
[tex]6=0+\frac{1}{2}\times 9.81\times t^2[/tex]
[tex]t=\sqrt{\frac{12}{9.81}}=1.106 s[/tex]
Thus time taken will also be 1.106 s in horizontal motion
[tex]R_x=u_xt+\frac{1}{2}at^2[/tex]
here a=0
[tex]R_x=15\times 1.106=16.59 m[/tex]
