Physics and Maths stuff

Apply Newton's 2nd law of motion (Force = mass*acceleration):

Vertical

Net force = -mg

-mg = m*d²y/dt²

so the mass cancels out and we are left with:

d²y/dt² = -g  which is the same as your equation Melody.

Horizontal

Net force = 0

0 = m*d²x/dt²

so dividing by the mass we are left with

d²x/dt² = 0

Mass is not relevant here.

.

According to your question, we are finding the height of second bounce.

If I say it's s, 

the u(initial velocity) of first bounce should be 0, because you said you are dropping, not adding force.

the v(final velocity) of first bounce should be u(initial velocity) of second bounce.

and the ball at peak of second bounce, the speed should be 0.

if we work out the acceleration by F=ma, (or if time or acceleration is given and you didn't mention) then we can work out s. By formula $${{\mathtt{v}}}^{{\mathtt{2}}} = {{\mathtt{u}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{as}}$$

$$0=(u_1) ^2+2as$$ (I'm setting peak of second bounce as end)

Hi radio,

That is a good question.  

I am not sure if the mass comes into this problem or not.  I don't think that it does.  

This is my guess.  I will ask someone more knowledgable to take a look at this question though.

I usually do these with calculus.

Initially. When the ball first leaves the ground

$$\\t=0, \\ \quad accel= \ddot y = -10m/s^2, \\ \quad vel=\dot y=u\;\;m/s,\\ \quad displacement=y=0\;metres\\\\
Ongoing\\
\ddot y = -10\qquad $Any stat point will be a maximum$\\
\dot y=-10t+u\\
y=-5t^2+ut\\\\$$

Maximum height will  be attained when velocity =0

$$\\-10t+u=0\\
10t=u\\
t=\frac{u}{10}\;\;seconds\\\\
$Now find y when $ t=\frac{u}{10}\\\\
y=-5\times \frac{u^2}{100}+u\times \frac{u}{10}\\\\
y=\frac{-u^2}{20}+ \frac{2u^2}{20}\\\\
y= \frac{u^2}{20}\\$$

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LET me see if I can do the same thing using physics formulas.

Initial velocity =u

final velocity = v =0

accel =a = -10

find displacement s

the only formula with u,v,a and s is  [4]

$$\\v^2=u^2+2as\\
0=u^2+2*-10s\\
-u^2=-20s\\\\
s=\frac{u^2}{20}$$

I got the same answer both ways.  That is good anyway   

The ball will reach a height of     $${\frac{{{\mathtt{u}}}^{{\mathtt{2}}}}{{\mathtt{20}}}}$$   metres on the first bounce

I'd really like a physics person to discuss the relevance of the mass of the ball. Please.  

The mass is not needed here.

.

I was hoping you would discuss it a little more than that Alan.

Is there anything more that you can say - I mean, why is mass irrelevant?

Thanks you Alan, much appreciated.