A lead ball is dropped into a lake from a diving board 4.50 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.40 s after it is released. How deep is the lake?
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calulate speed after falling 4.5 m
(1/2) m v^2 = m g h
so
v = sqrt (2 g h)
then
v t = 4.4
now try the others
To determine the depth of the lake, we can use the equation of motion for an object falling freely under gravity:
h = h0 + (1/2)gt^2
Where:
h = height or depth
h0 = initial height or depth
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken
In this case, the initial height (h0) is the height of the diving board, which is 4.50 m. The time taken (t) is 4.40 s. We need to find the depth of the lake, so we'll set h as the unknown value.
Since the ball sinks to the bottom of the lake with a constant velocity, we can assume the displacement in the vertical direction is zero. This means that the total distance traveled downwards is equal to the initial height from the diving board.
Therefore,
h0 = h + h'
Where:
h' = distance traveled downwards or depth
Rearrange the equation to find h:
h = h0 - h'
Substitute the known values:
h = 4.50 m - 0 m
Therefore, the depth of the lake is 4.50 meters.