A ball is to be shot from level ground toward a wall at distance x(Fig.4-43a). Fig 4-43b shows the ycomponent ${\mathit{v}}_{\mathbf{y}}$of the ball’s velocity just as it would reach the wall, as a function of that distance x. The scaling is set by ${\mathit{v}}_{\mathbf{y}\mathbf{x}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{}$and ${\mathit{x}}_{\mathbf{s}}\mathbf{=}\mathbf{20}\mathbf{}\mathit{m}$.What is the launch angle?

For the given graph, the scale is set by $v_{yx}=5.0m/s$and $x_s=20m$.

First, we have to find the time taken to travel horizontal distance. With the help of thistime, we can find the vertical component of the initial velocity. And from the graph, we can find the final velocity of the ball. From the y-intercept of the graph, we can conclude about the horizontal component of the initial velocity. The horizontal component of velocity is constant throughout the motion as there is no force in the horizontal direction.Thus, from the given data and the graph, using kinematic equations we can find the angle of launch.

The first kinematic equation is,

$v=v_0+at$ …(i)

The second kinematic equation is,

$s=v_{0}t+\frac{1}{2}a{t}^2$ …(ii)

The horizontal component of velocity is,

$v_{ox}=v_o\cos \theta$ …(iii)

The vertical component of initial velocity is,

$v_{0y}={\mathrm{v}}_0\sin \theta$ …(iv)

Launch angle is

$\theta ={\tan }^{-1}\left(\frac{v_{0y}}{v_{0x}}\right)$ …(v)

Given that the range of motion is 20m, from the graph it is clear that the initial velocity of ball is 5m/s and the final velocity of ball is also 5m/s.

The second kinematic equation gives the horizontal component ofthe initial velocity.

$S=v_{0x}t+\frac{1}{2}a{t}^2$

role="math" localid="1660895569268" $x=v_{0x}t$

$t=\frac{x}{v_{0x}}$

If we put the equation of time in the formula for the vertical velocity component we get,

$$\begin{gathered} v_y=v_{0y}-gt \\ =v_{0y}-\frac{gx}{v_{0x}} \end{gathered}$$

We know that the vertical component of velocity is zero at maximum height.

Also, the slope of the graph = $-\frac{1}{2}$

So, comparing it to the equation of line,we conclude,

$\begin{array}{l}-\frac{g}{v_{0x}}=-\frac{1}{2}\\ v_{0x}=19.6m/s\end{array}$

From the y-intercept of the graph, we can say that$v_{0y}=5.0\mathrm{m}/\mathrm{s}$

Now, use the equation (iv) and the x and y component of the initial velocity, to calculate the angle of launch.

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{v_{0y}}{v_{0x}}\right) \\ ={\tan }^{-1}\left(\frac{5.0m/s}{19.6m/s}\right) \\ =14.31° \end{gathered}$$

Therefore, the launch angle is $14.31°$.