Advance Mechanics Question (1 Viewer)

Nktnet · Dec 26, 2018

A bag of salt was projected vertically upwards from a height of 1 and remains in the air for 6.1 s.

(a) What speed was the bag thrown at, and
(b) What was the distance the bag travelled before returning to the height of 1 m?

InteGrand · Dec 26, 2018

Nktnet said:
A bag of salt was projected vertically upwards from a height of 1 and remains in the air for 6.1 s.

(a) What speed was the bag thrown at, and
(b) What was the distance the bag travelled before returning to the height of 1 m?

$\noindent \textbf{Hints.}$

$\noindent Let $y(t)$ be the vertical displacement (in m) of the bag from the ground after time $t$ (in s) (so $y(0) = 1$).$

$\noindent \textbf{\color{blue}{(a)}} Let $v > 0$ be the speed the bag was thrown at. Then using the ``$\text{change in position} = ut + \frac{1}{2}at^{2}$'' formula, we have $y(t) - y(0) = vt - \frac{1}{2}gt^{2} \Rightarrow y(t) = 1 + vt - \frac{1}{2}gt^{2}$, since $y(0) = 1$ (where $g$ is the acceleration due to gravity, about $9.8 \text{ m s}^{-2}$). Since the bag remains in the air for 6.1 s, what is the value of $y(6.1)$? Once you figure this out, use the equation for $y(t)$ to obtain an equation for $v$ (subbing in $t = 6.1$), and solve for $v$.$

$\noindent \textbf{\color{blue}{(b)}} Suppose the maximum height reached by the bag is $H$ (in m). Then the total distance travelled by the bag before returning to the original height of 1 m is just $2(H-1)$. Therefore, once you figure out $H$, you can figure out the answer to this question (it is $2(H-1)$, in m).$

$\noindent To figure out the maximum height $H$, you just need to find the maximum value of the height function, which is $y(t)= 1 + vt - \frac{1}{2}gt^{2}$. Since we know the value of $v$ (assuming you have done part (a)), you should be able to find this, assuming you remember how to find the maximum value of a quadratic function like this. If you don't remember this, then you can work it out as follows: the velocity in the vertical direction as a function of time is $v_{y}(t) = v - gt$ (make sure you know why this is!). The maximum height is achieved when $v_{y}(t) = 0$. (In other words, the bag stops instantaneously when it reaches the top of its flight.) So find the value of $t$ that makes $v_{y}(t) = 0$, and then substitute this into the formula for $y(t)$ to get the maximum height $H$.$

Advance Mechanics Question (1 Viewer)

Nktnet

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InteGrand

Well-Known Member

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