Question

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and her takeoff point is 1.80 m above the pool.

(a) How long are her feet in the air?
  s
(b) What is her highest point above the board?
  m
(c) What is her velocity when her feet hit the water?
  m/s

Additional Materials

Answer 1

a) taking vertically upwards direction as +ve ,

initial velocity=3 m/s

distance travelled downward=-1.8 m

acceleration=-9.8 m/s^2, downwards

let time travelled be t.

then -1.8=3*t+0.5*(-9.8)*t^2

solving for t,

we get t=0.985 seconds

hence her feet are in the air for 0.985 seconds.

b)at the highest point , her speed is 0 m/s

then using final velocity^2-initial velocity^2=2*acceleration*distance

using initial velocity=3 m/s

acceleration=-9.8 m/s^2

final velocity=0

then distance=0.4591 m

c) total height from the ground when she has velocity 0=1.8+0.4591=2.2591 m

then using initial velocity=0, acceleration=9.8 m/s^2

final velocity^2-initial velocity^2=2*acceleration*distance

we get final velocity=6.654 m/s