Answer:
For 0.24 sec the person was in the air.
Explanation:
Given that,
Height = 1 m
Initial velocity = 3 m/s
We need to calculate the time
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Where, u = initial velocity
s = height
Put the value into the formula
[tex]1 =3\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]4.9t^2+3t-1=0[/tex]
[tex]t = 0.24\ sec[/tex]
On neglecting the negative value of time
Hence, For 0.24 sec the person was in the air.
