Question

A ball rolls off a horizontal table at 0.45 m/s and strikes the ground 0.20 m horizontally from the base of the table. How high is the table?

Answer 1

Remember that:

in that moment when the ball is at the edge of the ball it has the max of potential energy wp=mgh. but it comes to loose it when it strikies the ground and you have to use wk=(m*v^2)/2.

Answer 2

try to use v*^2/(2g)=h (high)

Answer 3

If that is correct equation I can tell you how I got it:D

Answer 4

I don't think the energy conservative theory can be applied here, because if it is,
then the equation should be as follows,

if the velocity at the table = v
velocity of the ball when it hits the ground =u
height of the table = h
$$\frac{1}{2}mv^2+mgh=\frac{1}{2}mu^2$$ since there are two unknowns finding h will not be simple,

So the easiest thing to do is, since there is no acceleration to horizontal direction,
use,
$$\begin{align*}
\rightarrow S&=ut+\frac{1}{2}at^2\\
d&=vt_0\\
t_0&=\frac{d}{v}
\end{align*}$$
since the time taken for the ball to traveled is found use the above equation vertically, to find the height,
$$\begin{align*}
\downarrow S&=ut+\frac{1}{2}at^2\\
h&=(0)t_0+\frac{1}{2}gt_0^2\\
&=\frac{gt_0^2}{2}
\end{align*}$$