A diver leaves the springboard with a vertical velocity of 5.7 m/s, a horizontal velocity of 1.2 m/s, and a height of 3.49 m. How long will it take her to return to the same height

Time to return to same height=1.1632 s

Explanation:

This is a projectile motion

ux=1.2 m/s ,

uy=5.7 m/s

Time to return to same height =t= 2uy/g =2*5.7 /9.8 =1.1632 sec

Horizontal range =uxt =1.2*1.1632 =1.3959 m

Horizontal velocity remains same. But Vertical velocity will change. When it reaches same height it will attain same velocity. But after that vertical velocity will change due to gravity and covers a distance of s= 3.49 m.

Final vertical velocity vy, then  vy2=uy2+2gS =5.72+2*9.8*3.49 =100.894 or vy=10.044 m/sec

Magnitude of velocity when it hits water is =sqrt(1.22 +10.0442) =10.116 m/sec

Time to return to same height=1.1632 s

double the distance = divide strength by 4

thermal energy is the answer