# potential energy

Let $t_1$ be the time it takes an object to fall $x$ feet. The kinetic energy of a ball of mass m dropped vertically $x$ feet is $E_1=\frac 12mv_1^2$, where $v_1=h'(t_1)$. Find the formula for $E_1$ in terms of $m$ and $x$.

so $h(t_1)=-16t^2+x$, but here I need $h'(t_1)$, and when I calculate the derivative my $x$ dissapeared and became a number, so I can't get how am I supposed to use $x$ in a formula, in order to find the formula for $E_1$ in terms of $m$ and $x$?

• what is the whole problem?,where came from this $h\left( t_{ 1 } \right) =-16{ t }^{ 2 }+x$ write here full work – haqnatural Jul 23 '15 at 21:40
• $h(t)=-16t^2+vt+h$ it comes from here. where $h=x$ – Sarah Jul 24 '15 at 12:04

Not sure I understood the data correctly, but as I see it:

$h(t)$ is function of height with respect to time. because we start at hight of $x$ , and the pace of the fall is $16t^2$ then we have $h(t) = x - 16t^2$.

if it is indeed the case so $h'(t) = -32t$ (which is the speed - negative because falling - the speed is ofcourse has nothing to do with the $x$). Now, because $x=16t^2$ where $h(t) = 0$ [when we hit the ground]. we can say: $8\sqrt{x} = 32t =-h'(t)$.

can you take it from here ?

A change in coordinate system might make this a little easier to look at.

Let the positive h direction be downward, the initial height be zero: $h(0)=0$, and the initial velocity be zero: $h'(0)=0$.

The height below the starting point as a function of time will then be $h(t)=16t^2$. The time that you refer to as $t_1$ will then be the time the object takes to fall a distance x: $h(t_1)=x$. If you solve for $t_1$ in terms of x and substitute into $h'(t)$ you can get the velocity at the time the object has fallen a distance of x.

Hope this helps.

Energy, potential plus kinetic, is conserved (constant).

The potential energy $E_1$ of a body mass $m$ converted into kinetic energy by falling through distance $x_1$ is

$$mgx_1$$

This is the work done by the constant force $mg$ through a distance $x_1$.