# Doubled velocity problem - alternative calculation [closed]

Working through Tipler/Mosca I got stucked at a very simple task.

A stone is thrown up vertically (1 dimensional problem) with an initial velocity v and it is reaching a height of h. What happens to the height if you double the initial velocity

The solution is : solve both an equation for v and 2v for h:

$h= \frac{v^2}{2g}$

$h(new)= \frac{(2v)^2}{2g}$

then divide h(new) by h. All variables will cancel out and you will receive a ratio between h(new) and h which is 4. So the answer is: Double initial velocity will cause 4times higher throw. I understand that but I have a question about my first attempt which failed badly:

Why is my attempt running into bad results:

$v=\sqrt{2g\Delta y} \,\,\,\,|*2$ I tried to double the velocity

$2v=2\sqrt{2g\Delta y} \>\>\>|(...)^2$

$4v^2=4*2g\Delta y =8g\Delta y$

$\frac{4v^2}{8g}=\Delta y$ And here I am ending up with useless results.

Can someone explain why my attempt was determinated to fail ?

To not violate homework law: I am trying to understand what was my error in my strategy to solve the problem. The problem is solved and I understand the technique.

$\Delta y$ denotes maximum height in initial case when the initial velocity is $v$. Now, that you have fixed $\Delta y$ you cannot treat it as a variable for different velocities because $\Delta y$ is $h_{max}$ (which is a variable) for a specific case when initial velocity is $v$ . The new maximum height is not equal to $\Delta y$.

It doesn't matter how you manipulate the velocity-$\Delta y$ equation, you will always end up with the same result because manipulating this equation will bring the same change to both side of the equation.

$\Delta y=\frac{v^2}{2g}=\frac{4v^2}{8g}=H_{max,initial}$

If you want the correct result, you need a general equation with variables, viz., $v$ and $h_{max}$.

$$2gh=V^2$$

From the given data, $h_{max}=\Delta y$ and $V=v$

$\therefore 2g\Delta y=v^2$ $\tag 1$

But on doubling velocity, $h_{max}=H_{new}\neq\Delta y$ and $V=2v$

$\therefore 2gH_{new} ={(2v)}^2$ $\tag 2$

Now you will get your desired result.