Question

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.4 m/s, and her takeoff point is 1.95 m above the pool.

A) For how long are her feet in the air, in seconds?

Answer 1

Given : v₀ = 4.4 m/s ; y = -1.95 m [ 1.95 m below the diving board which is reference point]

Solution:

Applying the equation of motion as :

y = y₀ + (v₀)t +(1/2)at²

Here y₀ = distance of the feet from the reference point at t =0.

i.e. y₀ = 0

v₀ = initial velocity of the feet = 4.4 m/s

a = -g = -9.81 m/s²

Hence , we have :

(-1.95 m)= 0 +(4.4 m/s)t +(1/2)(-9.81 m/s²)t²

-1.95 = 4.4 t - 4.905 t²

Or , 4.905 t² - 4.4 t -1.95 =0

Solving this quadratic equation we get :

t = -0.325, 1.22

Time can not be negative hence the answer is t = 1.22 s