Answer:
He was 0.4781 s in air.
Explanation:
This question is on an object launched horizontally with zero angle and at a height H.
The formula to apply is ;
[tex]T=\sqrt{\frac{2H}{g} }[/tex]
where ;
T=time of flight = ?
H= height of the object was launched = 1.12 m
T= √2*1.12/9.8
T=√0.22857
T= 0.4781 seconds
