# Thread: velocity of a falling ball help

1. ## velocity of a falling ball help

In our simple model of the fight of a ball, the height
y above the foor at time t is

$\displaystyle y = V_{n}t-\frac{1}{2}gt^{2}$

where g is the acceleration due to gravity and vn is the velocity immediately after the nth
bounce.

The maximum height of the ball is reached when dy/dt = 0
Find the time
tmax at which this maximum occurs and the maximum height
ymax
that the ball reaches.

I am really not sure how to do this, since I do not know what vn is?

$\displaystyle \frac{dy}{dt} = V_{n} -gt = 0$

$\displaystyle V_{n} = gt$

dont know how to work ourt $\displaystyle y_{max}$

and $\displaystyle t_{max}$

2. ## Re: velocity of a falling ball help

$\displaystyle t_{max}$ is the time when the derivative is zero, and you found the equation $\displaystyle V_n=gt$ for that. So you solve for t and get $\displaystyle t_{max}=\frac{V_n}{g}$. Then you plug in to $\displaystyle y = V_{n}t-\frac{1}{2}gt^{2}$ to get the height at time $\displaystyle t_{max}$, which is $\displaystyle y_{max}$.

- Hollywood