Answer:
Explanation:
Given
maximum height=0.380 m
initial velocity=[tex]v_0[/tex]
[tex]H_{max}=\frac{v_0^2}{2g}[/tex]
[tex]0.380=\frac{v_0^2}{2\times 9.81}[/tex]
[tex]v_0=2.73 m/s[/tex]
The time of flight will be [tex]t=\frac{2v_0}{g}[/tex]
time to reach top +time to reach bottom will be same
[tex]t=\frac{2\times 2.73}{9.81}[/tex]
t=0.556 s