A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes time $0.420sec(s)$ to pass this window, which is of height $1.90meters(m)$. How far is the top of the window below the windowsill from which the flowerpot fell?
So what I have done is the following:
Since the pot is falling downwards, $a=-9.81\frac{m}{s^2}$.
The initial velocity is $v_0=0 \frac{m}{s}$ and the initial position is $x_0=0 m$.
Let the displacement of the x be $\Delta x=(1.90-y)m -0m$, where y is the distance between the top of the lower window and the bottom of the upper window. So, $x_f=(1.90+y)m$
Thus, I use the following formula to compute my answer:
$x_f=x_0 + v_0 t + \frac{1}{2}at^2 $
The goal is to find y.
$(1.90+y)m=(1.90-y)m + 0 \frac{m}{s} (0.420s) + \frac{1}{2} (9.81 \frac{m}{s^2})(0.420s)^2$
$(y)m=(-y)m + \frac{1}{2}(9.81)(0.420)^2m= [-y+\frac{1}{2}(9.81)(0.420)^2]m$
$y=-y+\frac{1}{2}(9.81)(0.420)^2$
$2y=\frac{1}{2}(9.81)(0.420)^2$
$y=\frac{1}{4}(9.81)(0.420)^2=1.03005$
Is my work correct because according to masteringphysics it isnt?
Best Answer
Assume upwards is positive. We have two equations:
$$\begin{align} -y =& \frac{1}{2}(-9.81)t^2\\ -(y+1.90) =& \frac{1}{2}(-9.81)(t+.420)^2 \end{align}$$
where $t$ is the time until the flowerpot reaches the top of the lower window. You should be able to find out this $t$ and then get $y$.