Answer to Question #94716 in Classical Mechanics for Jason

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 12.6 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

First, the time it takes to reach the water is calculated.

The vertical position is given by:

$Y=Y_{o}+V{0y}*t+\frac{1}{2}a_{y}t^{2}$

Where:

• Initial position in meters. $Y_{o}=12.6cm*\frac{1m}{100cm}=0.126m$
• Final position $Y=0m$
• Initial velocity $V_{oy}=0\frac{m}{s}$
• Vertical acceleration a$a_{y}=-9.8\frac{m}{s^{2}}$

Simplifying the expression.

$0=Y_{o}+\frac{1}{2}a_{y}t^{2}$

Expression for time

$0=Y_{o}+\frac{1}{2}a_{y}t^{2}\\ \frac{1}{2}a_{y}t^{2}=-Y_{o}\\t=\sqrt{\frac{-2Y_{o}}{a_{y}}}$

Evaluating numerically.

$t=\sqrt{\frac{-2*0.126m}{-9.8\frac{m}{s^{2}}}}=0.160s$

The velocity when it reaches the water is given by

$V_{y}=V_{oy}+a_{y}t$

Where:

• Velocity incial $V_{oy}=0\frac{m}{s}$
• velocity final $V_{y}=??$
• time $t=0160s$
• Gravity acceleration $a_{y}=-9.8\frac{m}{s^{2}}$

Numerically evaluating

$V_{y}=0\frac{m}{s}-9.8\frac{m}{s^{2}}*0.160s=-1.57\frac{m}{s}$

Upon entering the water moves at constant speed.

The distance traveled (depth) is given by:

$Y=V_{oy}*t$

Where

• Speed ​​(magnitude of velocity) $V_{oy}=1.57\frac{m}{s}$
• The given time is:$t=7.5s-0.160a=7.34s$ total time minus the time it takes to get to the water

Numerically evaluating

$Y=1.57\frac{m}{s}*7.34s=11.5m$

The depth of the lake is

$\boxed{Y=11.5m}$