Question

A cliff diver strikes the water with a velocity of 32 square root 5 feet per second. Find the height which the diver dove. V=square root 2gh

Answer 1

To find the height from which the diver dove, use the equation V = √(2gh), where V is the velocity of the diver, g is the acceleration due to gravity, and h is the height. The height from which the diver dove is 80 feet.

Step-by-step explanation:

To find the height from which the diver dove, we can use the equation V = √(2gh), where V is the velocity of the diver, g is the acceleration due to gravity (approximately 32 ft/s²), and h is the height. Given that V = 32√5 ft/s, we can solve for h.

V = √(2gh)

32√5 = √(2(32)(h))

Squaring both sides, we get:

(32√5)² = 64(32)(h)

1024(5) = 64(32)(h)

5120 = 64h

h = 80 ft

So, the height from which the diver dove is 80 feet.

Answer 2

80

Step-by-step explanation:

equation
`v=√(2gh)`

v =
`32√(5)` feet/s

g = 32 feet/s^2

plug in values

`32√(5)` =
`√(2*32*h)`

`\sqrt{32^(2)} * √(5) = \sqrt{32^(2)*5} = √(2*32*h)`

`32^2 * 5 = 2 * 32 * h`

32 * 5 = 2h

h = 32 * 5 / 2

h = 80