I threw a rock straight up with a speed of 13 m/s. I happened to be standing over a well when I threw it, so when instead of landing on the ground, it falls into the well. The surface of the water in the well is 5 meters below point when the rock left my hand. How long is it between when I released the rock and when it splashes on the water surface?

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Let's count a height from the level of the water surface. Then the height of the water surface is zero and the height of your hand is `H_0=5m.` Also we'll ignore air resistance.

The height `H` of a rock over the water surface is described by the formula

`H(t)=H_0+V_0t-g(t^2)/2`

where `V_0` is the initial speed. This is the projection on the upward axis, therefore `V_0` has plus sign and `g` has minus sign.

Our task is to find the time `t_1` at which `H(t_1)=0.` Of course `t_1` must be greater than zero.

 

In numbers, the equation is  `4.9t^2-13t-5=0,`

the positive solution is  `t_1=(13+sqrt(13^2+4*4.9*5))/(2*4.9) approx 3.0(s).`

This is the answer.

[The negative solution (with the minus before the root) has some sense. It corresponds to the imaginary situation when you would throw a rock from the water level. Then that time (its absolute value) is how long it would take for a rock to reach the height of your hand.]

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