Answer:11.875 m
Explanation:
Given
initial speed of wes ball([tex]u_1[/tex])=0 m/s
initial speed of Lindsay ball([tex]u_2[/tex])=21 m/s
let Height of building be h m
Consider vertical motion
[tex]h=u_1t+\frac{1}{2}g\left ( t+1.2\right )^2[/tex]
[tex]h=0+\frac{1}{2}g\left ( t+1.2\right )^2---------1[/tex]
here initial vertical velocity is zero and wes drop 1.2 s earlier than Lindsay
For Lindsay
[tex]h=u_2t+\frac{1}{2}gt^2[/tex]
[tex]h=21\times t+\frac{1}{2}gt^2---------------2[/tex]
Equating (1) & (2) we get
[tex]21\times t+\frac{1}{2}gt^2=\frac{1}{2}g\left ( t+1.2\right )^2[/tex]
42t=1.44g+2.4t
39.6t=1.44g
[tex]t=\frac{1.44\times 9.81}{39.6}=0.356 s[/tex]
Putting this value in (1)
[tex]h=\frac{1}{2}g\left ( 0.356+1.2\right )^2=11.875 m[/tex]
