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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 2 / Problem 2.83

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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QUESTION:

Visitors at an amusement park watch divers step off a platform 21.3 m above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). Air resistance may be ignored.

(a) Is the announcer correct in this claim?

(b) Is it possible for a diver to leap directly upward off the board so that, missing the board on the way down, she enters the water at 25 m/s? If so, what initial upward speed is required? Is the required initial speed physically attainable?

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QUESTION:

Visitors at an amusement park watch divers step off a platform 21.3 m above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). Air resistance may be ignored.

(a) Is the announcer correct in this claim?

(b) Is it possible for a diver to leap directly upward off the board so that, missing the board on the way down, she enters the water at 25 m/s? If so, what initial upward speed is required? Is the required initial speed physically attainable?

ANSWER:

Step 1 of 3

(a) Suppose the final velocity of the diver is $v$ . Now we know that the initial velocity $u=0$ m/s, the acceleration due to gravity is $g=9.8\ m/{s}^{2}$ and the distance travelled is $h=21.3\ m$ .

Now from the dynamics we know that

${v}^{2}={u}^{2}+2gh=0\ m/s+2gh=2gh$

$\Rightarrow{}v=\sqrt{2gh}=\sqrt{2(9.8\ m/{s}^{2})(21.3\ m)}=20.4$

Hence the velocity at which the diver will enter the water is 20.4 m/s.

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