Answer:
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
Velocity is define as the time rate taken for a change in acceleration
As it is going down the hill
