2.38 throw a glob of puddy straight up toward the ceiling

[karush]

you throw a glob of puddy straight up toward the ceiling.
which is 3.00 m above the point where the puddy leaves your hand.
the initial speed of the putty as it leaves your hand is 9.70 m/s

a. What is the speed of the puddy just before it strikes the ceiling?

$\begin{align*}\displaystyle
V^2 &= V_0^2 + 2ad \\
&=(9.7)^2 + 2(-9.8)(3)
&= 94.9-58.8\\
&= 36.1\\
V&\approx\color{red}{6 m/s}
\end{align*}$

b, How much time from when it leaves your and does it take the puddy to reach the ceiling?

\begin{align*}\displaystyle (6-9.7)/t&=-9.8\\
-3.7t&=-9.8s\\
t&=\frac{-9.8s}{-3.7}\\
t&\approx\color{red}{2.65s}
\end{align*}

OK no bk answer to this but think this is correct
also comment on format/process thanks

 

[MarkFL]

Neglecting drag, we know the acceleration of the projectile is given by:

\(\displaystyle \d{v}{t}=-g\)

Hence:

\(\displaystyle v(t)=\d{x}{t}=-gt+v_0\)

And:

\(\displaystyle x(t)=-\frac{g}{2}t^2+v_0t+x_0\)

We may write this as:

\(\displaystyle gt^2-2v_0t+2\Delta x=0\)

Using the quadratic formula (and discarding the larger root), we obtain:

\(\displaystyle t=\frac{v_0-\sqrt{v_0^2-2g\Delta x}}{g}\)

Hence:

\(\displaystyle v(\Delta x)=-g\left(\frac{v_0-\sqrt{v_0^2-2g\Delta x}}{g}\right)+v_0=\sqrt{v_0^2-2g\Delta x}\)

And so, plugging in the given data, we find:

\(\displaystyle v(3)=\sqrt{(9.7)^2-2(9.8)(3)}\approx5.94\frac{\text{m}}{\text{s}}\)

\(\displaystyle t=\frac{9.7-\sqrt{9.7^2-2(9.8)(3)}}{9.8}\approx0.38\text{ s}\)

 

[karush]

what an awesome answer