# What is the speed upon hitting the bottom in the scenario below?You stand at the base of a water fall. You decide to estimate the speed of the water as it hits the bottom. An information card...

What is the speed upon hitting the bottom in the scenario below?

You stand at the base of a water fall. You decide to estimate the speed of the water as it hits the bottom. An information card near by indicates that the speed of the river on the nearly horizontal ground before going over the falls is 16.4 m/s and the height of the falls is 52 m.

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Assuming that the water follows simple physics motion, then the height is:

`h=-1/2g t^2+v_0t+h_0`

where g=9.8 is the acceleration due to gravity, `v_0` is the initial velocity and `h_0` is the initial height.

The initial velocity of the waterfall is `v_0=-16.4` and the initial height is `h_0=52` . The final height is `h=0` . These can be substituted to get:

`-1/2(9.8)t^2-16.4t+52=0` which is a quadratic in t

`4.9t^2+16.4t-52=0` use quadratic formula but only need positive root

`t={-16.4+sqrt{16.4^2-4(4.9)(-52)}}/{2(4.9)}`

`=1.9889`

The we now substitute this time into the velocity equation `v=-g t+v_0` to get:

`v=-(9.8)(1.9888)-16.4=-35.9`

**The veclocity at the bottom of the waterfall is approximately 35.9 m/s.**