How Deep Is the Lake After Dropping a Ball from 30 Meters?

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SUMMARY

The discussion focuses on calculating the depth of a lake after a heavy ball is dropped from a height of 30.0 meters. The ball impacts the water with a final velocity of 24.3 m/s, which is derived using the equation vf^2 = vi^2 + 2ad, where 'a' is the acceleration due to gravity (-9.81 m/s²). After hitting the water, the ball continues to sink at this constant velocity for 10 seconds, resulting in a total depth of 243 meters.

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  • Knowledge of gravitational acceleration (9.81 m/s²)
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A heavy ball is dropped into a lake from a height of 30.0m above the water. it hits the water with a certain velocity and continues to sink to the bottom of the lake at this same constant velocity. It reaches the bottom of the lake 10.0s after it was dropped. How deep is the lake.


And the solution i got:


vf^2=vi^2+2ad
vf^2=2(-9.81)(30.0)
vf=24.3m/s


d=vt
=(24.3)(10)
=24.3m deep
 
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24.3*10=243m
 

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