0
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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