A spring-gun projects a small rock from the ground with speed ${\mathit{v}}_{\mathbf{0}}$at an angle $\mathit{\theta }\mathbf{°}$above the ground. You have been asked to determine${\mathit{v}}_{\mathbf{0}}$ . From the way the spring-gun is constructed, you know that to a good approximation ${\mathit{v}}_{\mathbf{0}}$is independent of the launch angle. You go to a level, open field, select a launch angle, and measure the horizontal distance the rock travels. You use$\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{80}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ and ignore the small height of the end of the spring-gun’s barrel above the ground. Since your measurement includes some uncertainty in values measured for the launch angle and for the horizontal range, you repeat the measurement for several launch angles and obtain the results given in Fig. 3.76. You ignore air resistance because there is no wind and the rock is small and heavy. (a) Select a way to represent the data well as a straight line. (b) Use the slope of the best straight-line fit to your data from part (a) to calculate${\mathit{v}}_{\mathbf{0}}$. (c) When the launch angle is$\mathbf{36}\mathbf{.}\mathbf{9}\mathbf{°}$ , what maximum height above the ground does the rock reach?

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

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

According to the newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$

$v^2=u^2+2as$

Where,$s,v,u,t,a$ are displacement, final velocity, initial velocity, time, and acceleration.

For the vertical and horizontal motion, the velocity component will be$u\sin \theta \text{andcos}\theta$respectively.

$Gravity=9.8m/s^2$

$\theta =36.9°$

The question has a graph representation of the variety of the rocks in the terms of the function of angle on which the rock was launched. The rock will free fall because of no air resistance. Now from the range formula,

$R=\frac{v_0^2\sin \left(2\theta \right)}{g}$

By this formula it is clear that the line formed by this equation $R\text{versussin}\left(2{\theta }_0\right)with\text{slope}{v}_0^2/g$will be a straight line because it is following the straight-line formula that is .

Now we wanted to show the graph that be will following as figure,

Slope of the graph is about as shown in the part (a) by the given data, so the velocity will be,

$\begin{array}{c}m=\frac{v_0^2}{g}\\ 11\text{\hspace{0.17em}m}=\frac{v_0^2}{9.8{\text{\hspace{0.17em}m/s}}^2}\\ v_0=10.38{\text{\hspace{0.17em}m/s}}^2\end{array}$

This will be the initial velocity.

From the newton’s formula of motion, here the final velocity will be zero on the maximum height for verticle movement,

$\begin{array}{c}{v}^2={\left(v_0\sin \theta \right)}^2+2as\\ 0^2={(10.38\text{\hspace{0.17em}m}/\text{s}\times \text{sin36.9}°)}^2+2\times 9.8\text{\hspace{0.17em}m}/\text{s}\times s\\ s=1.98\text{\hspace{0.17em}m}\end{array}$

This is the maximum height above the ground.