i need help with free fall!?

Question

a shoe is dropped into a lake from 20 ft above the lake surface. upon hitting the water the speed of the shoe is reduced by half and the shoe then sinks at constant speed. if it reaches the bottom 4 seconds after it was dropped, how deep is the lake?

Answer 1

The shoe falls under Earth's gravity at 20 feet = 6.096 meters

Velocity of the shoe when it reaches the water surface:

V(surface) = sqrt(2gd) = sqrt[2(9.81)(6.096)] = 10.94 m/s

Therefore the V(water) = 0.5V(surface) = 5.47 m/s

The shoe drops at this constant speed for t = 4 seconds and then hits the bottom. Therefore, the lake is

d = v(water)t = (5.47)(4) = 21.87 meters deep

Answer 2

First you want to know the speed of the shoe right before it hits the water, so use the kinematic equation that has distance, initial velocity (which is 0), final velocity, and acceleration (9.80 m/s^2)

"Upon hitting the water, the speed of the shoe is reduced by half"

This means take the answer you got and divide by two.

"The shoe then sinks at a constant speed"

This means the shoe isn't accelerating in any way and it is sinking at the same speed it was going AFTER it hit the water.

Since there is no acceleration afterwards, and you know the shoe fell for four seconds, and you want the distance it fell, just use the kinematic equation with velocity, time, and distance. Solve for the distance.

Answer 3

Dropping from 22 ft at 32 ft/sec^-2 and using

v^2 = 2gh = 44x32 = 1408 ==> v = 37.5 ft/s

Since the speed is reduced by half you have 37.5/2= 18.8 ft/s when it starts sinking at constant speed. Thus Depth of lake = vt = 18.8x4=75.2 ft

In all we assumed the shoe had no initial velocity when it was first dropped.

Answer 4

frist we need to find the speed at which the shoe strikes the surface of water.
applying the formula
v^2-u^2=2as
we get v as 10.93076 m/s
given the speed is reduced to half,
there fore its sped just before sinking is equal to 5.468351 m/s
as it sinks with constant speed,
we can apply the formula,D=speed*time
that would give us the distance as 21.861527m
therefore the lake is 21.861527m deep.

Answer 5

v^2 = 2as
v^2 = 2(32.2)(20)
v^2 = 1288 ft^2/sec^2
v = 35.88872 ft/sec when the shoe hits the water
reduced by 1/2,
v = 17.94436 fps
d = (17.94436 fps)*(4 sec.)
d = 71.77743 ft ≈ 72 ft