Time until an object reaches the bottom of a lake.

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 29.7 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 7.60 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.)

I attempted to answer this by having the two kinematic equations each eachother, and then using the quadratic equation to solve, but this seems to not be the right way.

Answer 1

This problem consists of two legs with constant acceleration so we can use the the first kinematic equation

x = x₀ + v₀ t + (1/2) a t²

twice,

(1) y = h₀ - (1/2) g t² (above the water)

(2) y = h_L - v t, (below the water)

where h₀ is the height above the water h_L is the height of the lake. From (1) with y=0, we can find the time to hit the water T₁, and from (2) we can fine the time it takes to hit the bottom of the lake.

(3) T₁ = (2 h₀/g)^1/2

(4) T_{2 =} h_L/v

(5) T₁+T₂ = T_total.

At this point we have 3 equations, and 4 unknowns. Although not stated we could assume the constant speed starts when the lure hits the water making,

(6) v² = v₀² + 2 a Δx = 2 g h₀ .

Putting (6) into 4 gives a solvable system of equations.

(3) T₁ = (2 h₀/g)^1/2

(4) T_{2 =} h_L/(2 g h₀)^1/2

(5) T₁+T₂ = T_total.

Giving

h = T_total(2 g h₀)^1/2 - 2 h₀ ≈ 17.76 m