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2 years ago

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

1 answer:

igor_vitrenko [27]2 years ago

6 0

Answer:

a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

b) Her highest height above the board is 0.82 m

c) Her velocity when her feet hit the water is 7.16 m/s

Explanation:

t = Time taken

u = Initial velocity = 4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=4-9.81\times t\\\Rightarrow \frac{-4}{-9.81}=t\\\Rightarrow t=0.41\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4\times 0.41+\frac{1}{2}\times -9.81\times 0.41^2\\\Rightarrow s=0.82\ m$

b) Her highest height above the board is 0.82 m

Total height she would fall is 0.82+1.8 = 2.62 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.62=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.62\times 2}{9.81}}\\\Rightarrow t=0.73\ s$

a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 0.73\\\Rightarrow v=7.16\ m/s$

c) Her velocity when her feet hit the water is 7.16 m/s

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