Answer:
50.2 cm
Explanation:
We are given that
Height ,h=3.5 m
Initial horizontal velocity,[tex]u_x=15 m/s[/tex]
Time,t=0.32 s
We have to find the distance of ball from the ground at 0.32 s.
Initial vertical velocity,[tex]u_y=0[/tex]
[tex]s=u_yt+\frac{1}{2}gt^2[/tex]
Where [tex]g=9.8 m/s^[/tex]
[tex]s=0+\frac{1}{2}(9.8)(0.32)^2[/tex]
[tex]s=0.502 m[/tex]
[tex]s=0.502\times 100=50.2 cm[/tex]
