This week’s post has something to do with Physics (position, velocity and acceleration) but it is mostly about good old plain Algebra and about reading the problem carefully.
For starters, since this takes place here on Earth, the acceleration of an object due to gravity is \(\mathbf{-32}\) feet per second square: \(\mathbf{a\left(t\right) = -32}\).
Here’s the problem:
You are riding a hot-air balloon which is rising vertically with a velocity of \(\boldsymbol{16}\) feet second. You decide to release a sandbag at the very instant the balloon is \(\boldsymbol{64}\) feet above the ground. Neglecting air resistance, how many seconds after its release will the sandbag hit the ground and at what velocity? (Recall the position equation: \(\mathbf{s\left(t\right) = -16\,t^{2} + v_{0}t + s_{0}}\)).
When I read this problem my first impulse was to come up with a quick answer in my head to the first part of the question: \(\mathbf{s\left(t\right) = -16\,t^{2} + 64}\) and solving for \(\mathbf{t}\) we have the sandbag hitting the ground at \(\mathbf{t = 2}\) seconds, (choosing the positive value). But, this way of thinking is wrong because the balloon, with you and the sandbag in it, is rising at \(\mathbf{16}\) feet second. This means that the very moment you release the sandbag, the sandbag is rising relative to the ground !
This means we have to include the balloon’s vertical velocity in the position equation: \(\mathbf{s\left(t\right) = -16t^{2} + 16\,t + 64}\):
\begin{align*}
\text{The sandbag will hit the ground when } s\left(t\right) &= 0 \\
-16\,t^{2} + 16\,t + 64 &= 0 \\
-16 \left(t^{2} – t -4\right) &= 0 \\
\text{Using the quadratic formula we get: } t &= \frac{1 + \sqrt{17}}{2} \text{(choosing the positive value)} \\
\text{The sandbag will hit the ground in } &\approx 2.56 \text{ seconds}
\end{align*}
Second part of the question: at what velocity will the sandbag hit the ground? The velocity function is the derivative of the position function: \(\mathbf{v\left(t\right) = s’\left(t\right) = -32\,t + 16}\).
We ask, then, what is the velocity at time \(\mathbf{\frac{1 + \sqrt{17}}{2}}\)?
\begin{align*}
v\left(\frac{1 + \sqrt{17}}{2}\right) &= -32\left(\frac{1 + \sqrt{17}}{2}\right) + 16 \\
&= -16 \left(1 + \sqrt{17}\right) + 16 \\
&= -16 \sqrt{17} \approx -65.97 \text{ feet per second}
\end{align*}
One final comment: we could have used \(\mathbf{v\left(2.56\right) = -32\left(2.56\right) + 16}\) giving us \(\mathbf{\approx -65.92}\) feet per second. Given the context of this problem, even an answer of “66 feet per second” is accurate enough. However, as a matter of principle, it is better to use exact numbers when possible, leaving any rounding for the end.
Quiz #2
We have the position function fresh in the mind, so let’s make use of it again.
Neglecting air resistance, with what initial velocity must an object be launched upward from a height of \(\mathbf{2\, meters}\) to reach a maximum height of \(\mathbf{200\, meters}\)? The problem is asked in meters, so the acceleration due to gravity is \(\mathbf{-9.8}\) meters per second square.
\begin{align*}
\text{Acceleration function: } a\left(t\right) &= -9.8 \\
\text{Velocity function: } v\left(t\right) &= -9.8\,t + v_{0} \\
\text{Position function: } s\left(t\right) &= -4.9\,t^{2} + v_{0}\,t + 2
\end{align*}
Answer next week.
Answer to Quiz #1
We had the following equation in last week’s post: \(\mathbf{\|\overrightarrow{\mathbf{BA}}\| = \frac{600}{\sin 15°} \approx 2318}\) lbs. The question was, how to arrive at an EXACT result using only pencil and paper?
There are common trig angles and trig identities that should be memorized, such as the values of \(\boldsymbol{30° \text{ and } 45°}\) angles, and the Sum and Difference Identity: \(\mathbf{\sin\left(a \pm b\right) = \sin\left(a\right) \cos\left(b\right) \pm \cos\left(a\right) \sin\left(b\right)}\). We proceed as follows:
\begin{align*}
\text{First, rewrite the equation as: } \|\overrightarrow{\mathbf{BA}}\| &= \frac{600}{\sin\left(45° – 30°\right)} \\
\text{Working aside with the denominator: } \sin\left(45° – 30°\right) &= \sin\left(45°\right) \cos\left(30°\right)\, – \cos\left(45°\right) \sin\left(30°\right) \\
&= \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right)\, – \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right) \\
\text{We obtain: } &= \frac{\sqrt{6}\, – \sqrt{2}}{4} \\
\text{The quotient then becomes: } &= \frac{600}{\frac{\sqrt{6}\, – \sqrt{2}}{4}} = \frac{2400}{\sqrt{6}\, – \sqrt{2}} \\
\text{We could go further and rationalize the denominator: } \frac{2400}{\left(\sqrt{6}\, – \sqrt{2}\right)} \cdot \frac{\left(\sqrt{6} + \sqrt{2}\right)}{\left(\sqrt{6} + \sqrt{2}\right)} &= \frac{2400\,\left(\sqrt{6} + \sqrt{2}\right)}{4} = 600\,\left(\sqrt{6} + \sqrt{2}\right)
\end{align*}
Thus arriving at an exact solution without using a graphing utility. By the way, I am learning that there are esoteric ways of reducing the sum \(\mathbf{\sqrt{6} + \sqrt{2}}\) to a single radical, but that is beyond the scope of my present knowledge.
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected]