Question

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.20 m above the pool. (a) How long are her feet in the air? .3 Incorrect: Your answer is incorrect. s (b) What is her highest point above the board? m (c) What is her velocity when her feet hit the water? m/s

Answer 1

The height of the swimmer's feet in the air at time $t$ is given according to

$y=1.20\,\mathrm m+\left(4.00\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g$ is the magnitude of the acceleration due to gravity (taken here to be 9.80 m/s^2).

a. Solve for $t$ when $y=0$:

$1.20\,\mathrm m+\left(4.00\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies\boxed{t=1.05\,\mathrm s}$

(The other solution is negative; ignore it)

b. At her highest point $y_{\rm max}$, the swimmer has zero velocity, so

$-\left(4.00\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1.20\,\mathrm m)\implies\boxed{y_{\rm max}=2.02\,\mathrm m}$

c. Her velocity at time $t$ is

$v=4.00\dfrac{\rm m}{\rm s}-gt$

After 1.05 s in the air, her velocity will be

$v=4.00\dfrac{\rm m}{\rm s}-g(1.05\,\mathrm s)\implies\boxed{v=-6.29\dfrac{\rm m}{\rm s}}$