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.)
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Expert's answer
2019-09-18T09:49:06-0400
First, the time it takes to reach the water is calculated.
The vertical position is given by:
Y=Yo+V0y∗t+21ayt2
Where:
Initial position in meters. Yo=12.6cm∗100cm1m=0.126m
Final position Y=0m
Initial velocity Voy=0sm
Vertical acceleration aay=−9.8s2m
Simplifying the expression.
0=Yo+21ayt2
Expression for time
0=Yo+21ayt221ayt2=−Yot=ay−2Yo
Evaluating numerically.
t=−9.8s2m−2∗0.126m=0.160s
The velocity when it reaches the water is given by
Vy=Voy+ayt
Where:
Velocity incial Voy=0sm
velocity final Vy=??
time t=0160s
Gravity acceleration ay=−9.8s2m
Numerically evaluating
Vy=0sm−9.8s2m∗0.160s=−1.57sm
Upon entering the water moves at constant speed.
The distance traveled (depth) is given by:
Y=Voy∗t
Where
Speed (magnitude of velocity) Voy=1.57sm
The given time is:t=7.5s−0.160a=7.34s total time minus the time it takes to get to the water
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