Question

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?

Answer 1

it will take 1.14 s to touch the water

Step-by-step explanation:

As we know that diver jumps upwards with initial speed

`v_i = 4 m/s`

now he takes off from initial height

`h = 1.80 m`

now we know that

`h = vt + (1)/(2)gt^2`

`-1.80 = 4 t - 4.9 t^2`

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

`t = 1.14 s`