A lead ball is dropped in a lake from a diving board 6.45 m above the water. It

Question

A lead ball is dropped in a lake from a diving board 6.45 m above the water. It hits the water with a certain velocity and then sinks to the bottom with the same constant velocity. It reaches the bottom 4.26 s after it is dropped. (a) How deep is the lake? (b) What is the magnitude of the average velocity of the ball for the entire fall? (c) Suppose the water is drained from the lake. The ball is now thrown from the diving board so that it again reaches the bottom in 4.26 s. What is the magnitude of the initial velocity of the ball?

Explanation / Answer

If the ball is dropped from height 5.12 (m), it's speed at the surface is given by: mv^2/2 = mgh v = sqrt(2gh) = sqrt(2*9.8*5.12) a) If the speed remains the same, and it reaches the bottom in 4.96 (s), the depth must be D = v*t = 4.96*v b) The average velocity is given by the total distance divided by the total time. The total distance = 5.12 + D; the total time is the time of fall to the water surface (= sqrt(2*5.12/g)) = sqrt(2*5.12/9.8))) plus 4.96. c) You can get the new total height as 5.12 + D, you can find how much time it takes to fall this distance, based upon initial velocity. When you set it equal to 4.96 seconds, you can invert the equation to find the initial velocity.

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