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A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. How long, in seconds, are her feet in the air?

Respuesta :

Answer:

it will take 1.14 s to touch the water

Explanation:

As we know that diver jumps upwards with initial speed

[tex]v_i = 4 m/s[/tex]

now he takes off from initial height

[tex]h = 1.80 m[/tex]

now we know that

[tex]h = vt + \frac{1}{2}gt^2[/tex]

[tex]-1.80 = 4 t - 4.9 t^2[/tex]

now by solving above equation for time "t" we have

[tex]t = 1.14 s[/tex]

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