Answer:3.06 s
Explanation:
Given
Vertical component of velocity i.e. [tex]v\sin \theta =u_y[/tex]=15 m/s
ball land on the same height as it is launched i.e. vertical displacement is zero
[tex]h=u_yt+\frac{at^2}{2}[/tex]
[tex]0=15\times t-\frac{9.81\times t^2}{2}[/tex]
[tex]t=\frac{2\times 15}{9.8}[/tex]
[tex]t=3.06 s[/tex]