We use the kinematic equation for vertical motion,
[tex]s=ut+\frac{1}{2}gt^2[/tex]
Here, [tex]s=-30\ m[/tex], [tex]t=1.50\ s[/tex] and [tex]g=-9.8\ m/s^2[/tex].
Therefore, the initial velocity when the object is at height 30 m, from above equation
[tex]-30\ m= u(1.50\ s)+\frac{1}{2} (-9.8\ m/s^2)(1.50\ s)^2\\\\ u=\frac{-18.98\ m }{1.50\ s} =-12.65\ m/s[/tex]
Now the displacement before the reaching the height 30.0 m above the ground, we use kinematic equation
[tex]u^2=U^2+2gh[/tex]
Here, [tex]u=-12.65\ m/s[/tex],[tex]U=0[/tex] because freely falling object, released from rest and [tex]g= -9.8\ m/s^2[/tex] and we take height -h because downward falling.
Substituting these values in above equation, we get
[tex](-12.65\ m/s)^2=0+2(-9.8\ m/s^2)(-h)\\\\ h=\frac{160.023}{19.6 }=8.16\ m[/tex].
Therefore, the total distance the object travels during the fall, s+h= 30 m+ 8.16 =38.16 m.
