Answer :
Answer:
14.05m/s
Explanation:
We are given that
Horizontal speed of diver=[tex]v_x=1.2m/s[/tex]
Distance, y=-10 m
It is taken as negative because the diver goes downward.
There is no air resistance therefore, there is acceleration due to gravity .
We know that
[tex]g=-9.8 m/s^2[/tex]
Initial vertical velocity, [tex]u_y=0[/tex]
[tex]v^2-u^2=2as[/tex]
Using the formula
[tex]v^2_y-0=2\times (-9.8)(-10)[/tex]
[tex]v_y=\sqrt{2(-9.8)(-10)}[/tex]
[tex]v_y=14 m/s[/tex]
[tex]v=\sqrt{v^2_x+v^2_y}[/tex]
Using the formula
[tex]v=\sqrt{(1.2)^2+(14)^2}=14.05m/s[/tex]
Hence, the speed of diver before just striking the water=14.05m/s