Answer to Question #95443 in Classical Mechanics for PIRA

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 15.0 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 7.50 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.)

Answer 1

Total fall consists of the accelerated motion in the air and constant velocity motion in the water:

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

where $h$ - height of the lure, $H$ - depth of the lake

So:

$$H=\sqrt{2gh}(t-\sqrt{\frac{2h}{g}})$$

$$H=\sqrt{2*9.8*0.15}(7.5-\sqrt{\frac{2*0.15}{9.8}})=12.6 m$$

Answer: depth of the lake is 12.6 meters