Answer to Question #94536 in Classical Mechanics for Kevin

Question

On a beautiful day you decide to go fishing with your younger brother Ben. Ben has not been fishing before so he doesn't know how to properly cast out his line. He puts the lure 25.2 cm above the water before letting it drop straight down. The line accelerates until it hits the water then continues to the bottom of the lake at a constant speed. It takes the lure 6.50 s to reach the bottom of the lake from when it was released by Ben.

How deep is the lake? (Assume the lure accelerates at the free fall rate of 9.8 m/s2 until it hits the water.)


What must have been his initial speed coming off the trampoline?

Answer 1

The fall consists of two parts: free fall in the air and constant velocity fall in the water:

$t=t_{air}+t_{water}=\sqrt{\frac{2h}{g}}+\frac{d}{v}=\sqrt{\frac{2h}{g}}+\frac{d}{\sqrt{2gh}}$

So,

$d=\sqrt{2gh}(t-\sqrt{\frac{2h}{g}})=\sqrt{2*9.8*0.252}(6.5-\sqrt{\frac{2*0.252}{9.8}})=13.9m$

Answer: The depth is 13.9 m