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 1.90 m above the pool. What is her velocity (in m/s) when her feet hit the water?

1.14\:s

Explanation:

The known values are: y_0=1.80\:m,\:y=0\:m,\:a=-9.80\:m/s^2,\:v_0=4.00\:m/s

Part A

Based from the knowns, the formula most applicable to solve for the time is \Delta y=v_0t+\frac{1}{2}at^2. If we rearrange the formula by solving for t, and substitute the given values, we have

t=\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a}

t=\frac{-4.00\:m/s\pm \sqrt{\left(4.00\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.80\:m\right)}}{-9.80\:m/s^2}

t=1.14\:s

spectral lines lyman is one set

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.