A diver springs upward from a board that is 4.40 m above the water. At the instant she contacts the water her speed is 13.5 m/s

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A diver springs upward from a board that is 4.40 m above the water. At the instant she contacts the water her speed is 13.5 m/s

and her body makes an angle of 78.1 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

Physics

1 answer:

Yakvenalex [24] · Answer 1 · 2023-09-19T05:46:45+03:00

The diver has the initial velocity, both (a) magnitude is 9.8 m/s and (b) direction is 73.5°.

<h3>What is free falling?</h3>

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy(P.E)

U =mgh

When the object strikes the ground, all the potential energy converted into kinetic energy.

K.E = 1/2mv²

where v is the speed just before hitting the ground.

A diver springs upward from a board that is 4.40 m above the water. At the instant, she contacts the water her speed is 13.5 m/s and her body makes an angle of 78.1 ° with respect to the horizontal surface of the water.

(a)

From energy conservation principle, initial and final mechanical energy are equal.

1/2mu² + mgh = 1/2mv²

where, u is the initial velocity of the diver.

u = sq rt (v² - 2gh)

u = sq rt (13.5² - 2x9.81x4.4)

u = 9.798 m/s or 9.8 m/s

Thus, the velocity of the diver is 9.8 m/s.

(b)

The horizontal component of velocity will remain constant.

The horizontal component of acceleration is zero.

Then,