Question

A physics student tossed a ball vertically straight upward.

Answer 1

Answer:

Explanation: ns along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. (Figure) illustrates the notation for displacement, where \mathbf{s} is defined to be the total displacement and \mathbf{x} and \mathbf{y} are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation \mathbf{A} to represent a vector with components {\mathbf{A}}_{x} and {\mathbf{A}}_{y}. If we continued this format, we would call displacement \mathbf{s} with components {\mathbf{s}}_{x} and {\mathbf{s}}_{y}. However, to simplify the notation, we will simply represent the component vectors as \mathbf{x} and \mathbf{y}.)

Review of Kinematic Equations (constant a)

x={x}_{0}+\stackrel{-}{v}t

\stackrel{-}{v}=\frac{{v}_{0}+v}{2}

v={v}_{0}+\text{at}

x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}

{v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right)\text{.}

The total displacement \mathbf{s} of a soccer ball at a point along its path. The vector \mathbf{s} has components \mathbf{x} and \mathbf{y} along the horizontal and vertical axes. Its magnitude is s, and it makes an angle \theta with the horizontal.

A soccer player is kicking a soccer ball. The ball travels in a projectile motion and reaches a point whose vertical distance is y and horizontal distance is x. The displacement between the kicking point and the final point is s. The angle made by this displacement vector with x axis is theta.

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1.Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so {A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta and {A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta are used. The magnitude of the components of displacement \mathbf{s} along these axes are x and \mathrm{y.} The magnitudes of the components of the velocity \mathbf{v} are {v}_{x}=v\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta and {v}_{y}=v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\theta ,} where v is the magnitude of the velocity and \theta is its direction, as shown in (Figure). Initial values are denoted with a subscript 0, as usual.

Step 2.Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

\text{Horizontal Motion}\left({a}_{x}=0\right)

x={x}_{0}+{v}_{x}t

{v}_{x}={v}_{0x}={v}_{x}=\text{velocity is a constant.}

\text{Vertical Motion}\left(\text{assuming positive is up}\phantom{\rule{0.25em}{0ex}}{a}_{y}=-g=-9.\text{80}{\text{m/s}}^{2}\right)

y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t

{v}_{y}={v}_{0y}-\text{gt}

y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\mathrm{gt}}^{2}

{v}_{y}^{2}={v}_{0y}^{2}-2g\left(y-{y}_{0}\right)\text{.}

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4.Recombine the two motions to find the total displacement\mathbf{\text{s}} and velocity \mathbf{\text{v}}. Because the x – and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing A=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}} and \theta ={\text{tan}}^{-1}\left({A}_{y}/{A}_{x}\right) in the following form, where \theta is the direction of the displacement \mathbf{s} and {\theta }_{v} is the direction of the velocity \mathbf{v}:

Total displacement and velocity

s=\sqrt{{x}^{2}+{y}^{2}}

\theta ={\text{tan}}^{-1}\left(y/x\right)

v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}

{\theta }_{v}={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right)\text{.}

(a)

Answer 2

Answer:

c. The magnitude of vertical velocity is at a maximum value.

Explanation:

I think this is correct I'm not entirely sure though