Aball is dropped from a boat so that it strikes the surface of a lake with a speed of 16.5 ft/s. while in the water, the ball experiences an acceleration of a = 10 â 0.8v, where a and v are expressed in ft/s2 and ft/s, respectively. the ball takes 5.7 s to reach the bottom of the lake.

Use the surface of the water as a reference.
Measure distance, velocity, and acceleration as positive downward.

Define
t = time, s
h = distance, m
a = 10 m/s², acceleration
u = 16.5 ft/s, initial velocity

The velocity at time t is v = u + at
When the ball reaches the bottom of the lake at t = 5.7s, its velocity is
v = (16.5 ft/s) + (10 ft/s²)*(5.7 s) = 73.5 m/s

Distance traveled at time t is h = ut + (1/2)at².
The ball reaches the bottom of the lake at 5.7 s.
Therefore the depth of the lake is
H = (16.5 ft/s)*(5.7 s) + 0.5*(10 ft/s²)*(5.7 s)² = 256.5 m

The ball reaches the bottom of the lake with velocity 73.5 m/s
The depth of the lake is 256.5 m.

Milliseconds. . . . . .