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SSSSS [86.1K]
3 years ago
12

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?
Physics
1 answer:
igor_vitrenko [27]3 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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Read 2 more answers
A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
Citrus2011 [14]

Answer:

part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

  • velocity of the river due to east = v_r\ =\ 1.60\ m/s.
  • velocity of the boat due to the north = v_b\ =\ 10.3\ m/s.

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

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total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

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