Question

A swimmer bounces almost straight up from a diving board and falls vertically feet first into a pool.she starts with a speed of 4.4m/s and her take off point is 1.8m above the pools surface. a. for how long are her feet in the air in seconds
b. what is the height above the board in meters of her highest point.
c. what is her velocity in meters per second when her feet hit the water. Assume the vertical component of the velocity is is positive upwards

Answer 1

Answer:

a) 1.20227 seconds

b) 0.98674 m

c) 7.3942875 m/s

Explanation:

t = Time taken

u = Initial velocity = 4.4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=4.4-9.81\times t\\\Rightarrow \frac{-4.4}{-9.81}=t\\\Rightarrow t=0.44852\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.4\times 0.44852+\frac{1}{2}\times -9.81\times 0.44852^2\\\Rightarrow s=0.98674\ m$

b) Her highest height above the board is 0.98674 m

Total height she would fall is 0.98674+1.8 = 2.78674 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.78674=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.78674\times 2}{9.81}}\\\Rightarrow t=0.75375\ s$

a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 0.75375\\\Rightarrow v=7.3942875\ m/s$

c) Her velocity when her feet hit the water is 7.3942875 m/s