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maw
2 months ago
7

[High Dive) 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 r

esistance may be ignored in this problem.) ve] Visitors at an amusement park watch divers step off a platform 70 ft (a) Is the announcer correct in this claim? Please explain. (b) Is it possible for a diver to leap directly upward off the board so that, missing the platform on the way down, the diver enters the water at 25 m/s? If so, what initial upward speed is required? Is the required initial speed physically attainable? How do you know?
Physics
1 answer:
Yuliya22 [3.3K]2 months ago
3 0

Answer:

(a) The announcer's statement is inaccurate since the divers enter the water with a speed of 20.4, not 25 m/s as claimed

(b)  a diver could enter the water at a speed of 25 m/s if their initial velocity was 14.4 m/s. However, achieving this upward initial speed is not physically feasible

Explanation:

(a)

To determine the final velocity V_{f} for an object moving a distance h with an initial vertical velocity V_{i} the kinematic equation takes the form

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Using g for a, where g is taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

With an initial speed of zero, we find the final velocity using substitutions, noting that 70 ft converts to 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Thus, divers descend at a rate of 20.4 m/s

The announcer's statement is incorrect as divers enter at 20.4 and not at 25 m/s as stated

(b)

Divers can achieve a velocity of 25 m/s if they possess an initial velocity. Utilizing the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

With a final speed of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

This indicates that it is conceivable for a diver to enter the water at 25 m/s assuming an initial velocity of 14.4 m/s

Ultimately, the required initial velocity is not achievable physically

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Answer:

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Explanation:

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