Q77P - University Physics with Modern Physics

Question

You have constructed a hair-spray-powered potato gun and want to find the muzzle speed ${\mathit{v}}_{\mathbf{o}}$of the potatoes, the speed they have as they leave the end of the gun barrel. You use the same amount of hair spray each time you fire the gun, and you have confirmed by repeated firings at the same height that the muzzle speed is approximately the same for each firing. You climb on a microwave relay tower (with permission, of course) to launch the potatoes horizontally from different heights above the ground. Your friend measures the height of the gun barrel above the ground and the range R of each potato. You obtain the following data: 

 

Launch height h	Horizontal range R
$\begin{array}{l}\begin{array}{l}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{m}\\ \mathbf{6}\mathbf{.}\mathbf{00}\mathbf{m}\\ \mathbf{9}\mathbf{.}\mathbf{00}\mathbf{m}\end{array}\\ \mathbf{12}\mathbf{.}\mathbf{00}\mathbf{m}\end{array}$	$\begin{array}{l}\begin{array}{l}\mathbf{10}.\mathbf{4}\mathbf{m}\\ \mathbf{17}.\mathbf{1}\mathbf{m}\\ \mathbf{21}.\mathbf{3}\mathbf{m}\end{array}\\ \mathbf{25}.\mathbf{8}\mathbf{m}\end{array}$

 

Each of the values of h and R has some measurement error: The muzzle speed is not precisely the same each time, and the barrel isn’t precisely horizontal. So you use all of the measurements to get the best estimate of ${\mathit{v}}_{\mathbf{o}}$. No wind is blowing, so you decide to ignore air resistance. You use$\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{8}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ in your analysis. (a) Select a way to represent the data well as a straight line. (b) Use the slope of the best-fit line from part (a) to calculate the average value of${\mathit{v}}_{\mathbf{o}}$ . (c) What would be the horizontal range of a potato that is fired from ground level at an angle of $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$above the horizontal? Use the value of ${\mathit{v}}_{\mathbf{o}}$that you calculated in part (b).

Step-by-Step Solution

Answer

(a) Here graph will be a straight line for${\text{R}}^{\text{2}}\text{ versus h and slope will be}\frac{{\text{2v}}_{\text{o}}^{\text{2}}}{\text{g}}$

.

(b) Average velocity will be$v_o=16.48\text{m}/\text{s}$ .

 

(c)Horizontal range of potato will be$R=\text{24.01m}$ .

Step 1: Newton’s law of motion

The velocity contains two components, one is a horizontal component and another one is vertical. 

 

The newton’s laws of motion show the relationship between the moving particle or object’s displacement, initial and final velocity, acceleration, and time.

 

According to the newton’s laws of motion,

 $\mathit{s}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$

Where,s ,$u$ , $t$, and are displacement, initial velocity, time, and acceleration respectively

Step 2: Given

Gravity$g=9.8\text{m}/{\text{s}}^{\text{2}}$

 

Angle above the horizontal =$30.0°$

 

Launch height h	Horizontal range R
$\begin{array}{l}\begin{array}{l}2.00m\\ 6.00m\\ 9.00m\end{array}\\ 12.00m\end{array}$	$\begin{array}{l}\begin{array}{l}10.4m\\ 17.1m\\ 21.3m\end{array}\\ 25.8m\end{array}$

Step 3: (a) Given data as a straight line

Here launch height or displacement is h, horizontal range is R, gravity is g, $v_o$is muzzle velocity of potatoes.

 

For the vertical motion, the initial velocity is zero so launch height or displacement h will be,

                                                                                         

$\begin{array}{c}h=\frac{1}{2}g{t}^2\\ t=\sqrt{2h/g}\\ R=v_{o}t\\ R^2=\frac{2v_o^2}{g}h\end{array}$

 

Here graph will be a straight line for ${\text{R}}^{\text{2}}\text{ versus h and slope will be}\frac{{\text{2v}}_{\text{o}}^{\text{2}}}{\text{g}}$.

Step 4: (b) Average value of v o

From the given table will put the value of R and h here, g value is also given in the question.

 

$\begin{array}{c}{R}^2=\frac{2v_o^2}{g}h\\ {\left(25.8\right)}^2=\frac{2v_o^2}{\text{9.8m}/{\text{s}}^{\text{2}}}\times 12.00\\ v_o=16.48\text{m}/\text{s}\end{array}$

 

Average velocity will be$v_o=16.48\text{m}/\text{s}$ .

Step 5: (c) Horizontal range of potato

The horizontal range of the potato will be, 

 

$v_o=16.48\text{m}/\text{s}$   and angle is$30.0°$

 

$\begin{array}{c}R=\frac{v_o^2\sin 2\theta }{g}\\ R=\frac{{(16.48\text{m}/\text{s})}^2\times \sin 2\times 30°}{\text{9.8m}/{\text{s}}^{\text{2}}}\\ R=\text{24.01m}\end{array}$

 

Horizontal range of potato will be$R=\text{24.01m}$ .