A lead ball is dropped into a lake from a diving board 5.24 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 4.86 s after it is dropped. (Assume the positive direction is upward.)

Question

A lead ball is dropped into a lake from a diving board 5.24 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 4.86 s after it is dropped. (Assume the positive direction is upward.) 
(a) How deep is the lake?

(b) What is the average velocity of the ball?

(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 4.86 s. What is the initial velocity of the ball?

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Jajsjs · Accepted Answer

Sjsjsjd

Damon · Answer

first the fall from the board to the surface of the lake (call water level z = 0
0 = 5.24 - (1/2)g t^2
5.24 = 4.9 t^2
t = 1.03 s
v = 0 - 9.8t
v = - 10.0 m/s when it hits
z = -10 m/s * (4.86-1.03)
z = -38.3 = depth of lake
The ball went (5.24+38.3)meters down in 4.86 seconds so
average velocity = -8.95 m/s
with no water
-38.3 = 5.24 + Vo t - 4.9 t^2
-43.54 = Vo(4.86) - 4.9 (4.86)^2
4.86 Vo = 71.87
Vo = + 14.8 m/s UP