Question

An Olympic diver is on a diving platform 8.40 m above the water. To start her dive, she runs off of the platform with a speed of 1.2 m/s in the horizontal direction. What is the diver speed, in m/s, just before she enters the water?m/s

Answer 1

Given data

*The given distance above the water is s = 8.40 m

*The given initial speed of the Olympic diver is u = 1.2 m/s

*The value of the acceleration due to gravity is a = 9.8 m/s^2

The formula for the diver speed, in m/s, just before she enters the water is given by the equation of motion as

`v^2=u^2+2as`

Substitute the known values in the above expression as

`\begin{gathered} v^2=(1.2)^2+2*9.8*8.40 \\ v=12.88\text{ m/s} \end{gathered}`

Hence, the diver speed, in m/s, just before she enters the water is v = 12.88 m/s