gAn Olympic diver is on a diving platform 5.40 m above the water. To start her dive, she runs off of the platform with a speed of 1.17 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water

The diver's speed, in m/s, just before she enters the water = 10.19 m/s

Explanation:

Assuming the horizontal velocity of the diver remains constant by neglecting the air resistance.

V^2 = U^2 + 2as

Where, V^2 = the diver's final velocity before impact with the water

            U^2 = Initial diver's velocity as she leaves the diving platform

                a = acceleration due to gravity

                s = the displacement

V =  √U^2 + √2as

= √(5.40 m/s)^2 + √2(9.81 m/s/s) (1.17 m)  

= √ 29.16 + √ 22.9554

=  5.40 + 4.79

= 10.19 m/s