#### Answer

$d=v_{x}t=16,86\dfrac {m}{s}\times 4s=67.44m$

#### Work Step by Step

Lets play it backwards and assume ball is launched from the top of building at point A with an angle of $\theta$ and it reaches the ground after $t=4s$ and the height of building is $h=20m$ So lets write the equation showing height of ball respect to ground at any given time. $v$ is the velocity of the ball at point A $H=h+vt\sin \theta -g\dfrac {t^{2}}{2}$ When it reaches the ground $H=0$ so $0=h+vt\sin \theta -g\dfrac {t^{2}}{2}\Rightarrow 0=20m+v\times 4s\times\sin {60}-\dfrac {9.8\dfrac {m}{s^{2}}}{2}\left( 4s\right) ^{2}$ İf we solve the equation we get $v=16,86\dfrac {m}{s}$ So lets find horizontal component of velocity of the ball which doesnt change becouse there is no horizontal force acting on the ball $v_{x}=v\cos \theta =16,86\cos 60^{0}=8.43\dfrac {m}{s}$ So the horizontal distance ball travelled until it reached to the top of building will be $d=v_{x}t=16,86\dfrac {m}{s}\times 4s=67.44m$