2.38 throw a glob of puddy straight up toward the ceiling

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In summary, the puddy thrown straight up toward the ceiling with an initial speed of 9.70 m/s reaches a speed of 6 m/s just before it strikes the ceiling, and it takes approximately 2.65 seconds for the puddy to reach the ceiling from when it leaves your hand. Using the quadratic formula, we can also calculate the speed and time for the puddy at a specific distance, such as 3.00 m above the point where it leaves your hand.
  • #1
karush
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MHB
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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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  • #2
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:

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

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

FAQ: 2.38 throw a glob of puddy straight up toward the ceiling

1. What is the purpose of throwing a glob of puddy straight up toward the ceiling?

The purpose of this experiment is to study the effects of gravity on a falling object. By throwing the puddy straight up, we can observe how it moves and eventually falls back down to the ground.

2. How high will the puddy go when thrown straight up?

The height of the puddy's trajectory will depend on the initial velocity at which it is thrown and the force of gravity. However, it will eventually reach a maximum height before falling back down due to the force of gravity.

3. What factors might affect the puddy's trajectory when thrown straight up?

The main factors that can affect the puddy's trajectory are the initial velocity at which it is thrown, the force of gravity, and the air resistance. Other factors such as wind or humidity may also have an impact.

4. How can this experiment be used to understand physics concepts?

By studying the puddy's motion, we can apply principles of physics such as Newton's laws of motion and the concept of acceleration due to gravity. This experiment can also help us understand concepts such as potential and kinetic energy.

5. Can this experiment be replicated with other objects?

Yes, this experiment can be replicated with other objects such as balls, balloons, or even paper planes. As long as the object is thrown straight up, we can observe similar effects of gravity and motion. However, the results may vary depending on the object's weight and shape.

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