MHB 2.38 throw a glob of puddy straight up toward the ceiling

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The discussion focuses on the physics of a glob of putty thrown straight up towards a ceiling 3.00 m high with an initial speed of 9.70 m/s. The calculations reveal that the speed of the putty just before it strikes the ceiling is approximately 6 m/s. Additionally, it takes about 2.65 seconds for the putty to reach the ceiling. The participants validate their calculations using kinematic equations, confirming the results. Overall, the thread emphasizes the application of physics principles to solve projectile motion problems.
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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
 
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Neglecting drag, we know the acceleration of the projectile is given by:

$$\d{v}{t}=-g$$

Hence:

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

And:

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

We may write this as:

$$gt^2-2v_0t+2\Delta x=0$$

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

$$t=\frac{v_0-\sqrt{v_0^2-2g\Delta x}}{g}$$

Hence:

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

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

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

$$t=\frac{9.7-\sqrt{9.7^2-2(9.8)(3)}}{9.8}\approx0.38\text{ s}$$
 
what an awesome answer:cool:
 
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