Answer:
1.25 s
Explanation:
We are given that
Initial velocity of boy=u=0
Time=0.883 s
Let x be the total distance covered by boy in entire trip.
s=[tex]\frac{x}{2}[/tex]
We know that
[tex]s=ut+\frac{1}{2}gt^2[/tex]
Where [tex]g=10m/s^2[/tex]
Substitute the values
[tex]\frac{x}{2}=0\times 0.883+\frac{1}{2}(10)(0.883)^2[/tex]
[tex]\frac{x}{2}=3.898[/tex]
[tex]x=3.898\times 2=7.796[/tex]m
Again using the above formula
[tex]7.796=0(t)+\frac{1}{2}(10)t^2[/tex]
[tex]7.796=5t^2[/tex]
[tex]t^2=\frac{7.796}{5}=1.5592[/tex]
[tex]t=\sqrt{1.5592}=1.25 s[/tex]
Hence, 1.25 s passes during his entire tripe from the top down to the water.
