Posted by Ray on Monday, September 7, 2009 at 6:21pm.
A lead ball is dropped into a lake from a diving board 5.42 m above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom 4.98 s after it is dropped. (Assume the positive direction is upward.)
(a) How deep is the lake?
1Your answer is incorrect. m
(b) What is the average velocity of the ball?
2Your answer is incorrect. m/s
(c) Suppose that all the water is drained from the lake. The ball is now thrown from the diving board so that it again reaches the bottom in 4.98 s. What is the initial velocity of the ball?

Physics  MathMate, Tuesday, September 8, 2009 at 8:22am
Consider the free fall height of S=5.42 m.
Initial velocity u=0 m/s.
Final velocity = v.
Acceleration due to gravity, g=9.81 m/s/s
Use the relation
v²u² = 2aS
to get
v=sqrt(2*(9.81)*(5.42)+0)
=10.312 m/s
The time required for the drop (to the surface of the lake)
= 10.312/9.81 m/s / m/s/s
= 1.051 seconds
There is enough information to complete the calculations. Post your work any time if you encounter problems.
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