Question

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 5.00 m/s, and her takeoff point is 1.70 m above the pool. (a)How long (in s) are her feet in the air? (b)What is her highest point (in m) above the board? (c) What is her velocity (in m/s) when her feet hit the water?

Answer 1

The motion is governed by equations:

y = ut + 1/2at²

v = u + at

Let upwards direction is positive and downwards is negative.

(a) Taking u = 5 m/s and acceleration g = -9.8 m/s²

as final distance covered in total time is -1.7 m

-1.7 = 5t - 0.5 x 9.8t²

4.9t² - 5t - 1.7 = 0

Which gives t = 1.29 s positive root of quadratic eqn.

Hence feet are in air for 1.29 s

(b) At highest point

mgh = 1/2mv²

h = v²/2g = 25/2 x 9.8 = 1.27 m from initial height

Hence net height will be 1.7 + 1.27 = 2.97 m

(c) using v = u + at

And t = 1.29 s

v = 5 - 9.8 x 1.29 =- 7.642 m/s (downwards)