Speed under water?

Question

A person jumped off a diving board 4.4m above the water's surface into a deep pool. The person's downward motion stops 1.8m below the surface of the water. Estimate the average deceleration (magnitude) of the person while under water

HUH?

Answer 1

First find out how fast the diver is going when they reach the water.  Use



  y = (v^2 - v0^2)/(2a)  

v = speed at water, v0 = speed at start of dive = 0, a  = g = 9.8 m/s^2 and y = 4.4 m

y = v^2/(2g)  ---> v = sqrt(2gy)  (you can plug numbers in)

Now you are told the diver stops 1.8m below the surface of the water and you know the speed they started with so you cna use the same equation to find acceleration:

y1= (v^2 - v0^2)/(2a)  where now v = 0 and v0 = sqrt(2gy)

y1 = -v0^2/(2a)  ---> a = -v0^2/(2y1)

a = -2gy/(2y1) = -g*(y/y1)

Substitute y = 4.4, y1 = 1.8 and g = 9.8 and you have your answer.  Note the "-" sign.  It is there because the acceleration is in the opposite direction of gravity