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Considering the ground reference.

The following equation is employed

Y=Y0+Voy*t+(1/2)*ay*t^2

where:

initial position

Y0=782m

final position

Y=0m

initial velocity. is dropped.

Voy=0m/s

acceleration of gravity

ay=-9.81m/s^2

eliminating null terms

0=Y0+(1/2)*ay*t^2

Expression of time

t=square root of(-(2*Y0)/ay)=square root of(-(2*782m)/-9.81m/s^2)=12.67s

finally

t=12.67s

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