Time Time

Yeah the answer is wrong

I have updated the answer to $\frac{2}{n-1} \sqrt{ \frac{2x}{g} }$ .

During the time interval of $2\sqrt{\frac{2x}{g}}$ the juggler has to throw n balls, not n-1 balls.

This is my first problem and solution so i heartily ask for any inconvenience or mistake.

Please correct the answer..... it says your solution is wrong..(while its correct) :)

where did you get the 2 on the outside of the square root?

For those who are wondering where the $T = \sqrt{\frac{2x}{g}}$ came from, here is the derivation.

We start out with the simple equation for (vertical for our purposes) the acceleration near the surface of the earth for a point mass.

$a(t) = -g$

Then we integrate that twice to get the linear displacement function. Remember the initial conditions are, generally, $v(0) = v_0$ and $d(0) = d_0$ .

$d(t) = \int (\int a(t) dt) dt = \frac{-g}{2}t^2 + v_0t + d_0$

Where $v_0$ is the initial velocity, and $d_0$ is the initial displacement.

Now in the case of the point mass merely dropping off a certain height $x$ and reaching ground level ( $d(t)=0$ ), we reduce the equation to.

$d(t) = 0 = \frac{-g}{2}t^2 + (0)t + x$

Now if we just move the $x$ to the left side, put the $2$ and the $g$ on the left side in their proper places, and then square root the whole thing, we get

$t = \sqrt{\frac{2x}{g}}$

Which is the time taken for the point mass to fall from a certain height $x$ and reach 'ground' level.

Simple is using the 2nd law of motion, Without using the derivatives to show that T=..... H=ut+1/2gt sq. U=0m/s,t=sq. Root of 2H/g Here, x=H Hence the required solution.