Answer:
1.03s
Given, acceleration a=−g=−9.8m/s
2
(minus sign for motion against gravity); final velocity (top most point) is v=0m/s
Vertical distance traveled d=1.29m
If u be the initial velocity, using formula v
2
−u
2
=2ad
0
2
−u
2
=2(−9.8)(1.29)
⟹ u=5.03m/s
If t be the time to peak.
Using v=u+at
0=5.03−(9.8)t
We get t=5.03/9.8=0.513s
So, the hang time will be double of peak time i.e. t
hang
=2×0.513=1.03s
