Question

Figure 4-23 shows three situations in which identical projectiles are launched (at the same level) at identical initial speeds and angles. The projectiles do not land on the same terrain, however. Rank the situations according to the final speeds of the projectiles just before they land, greatest first.

Short answer

Rank: a > b > c

Step-by-step solution

Given information

$a=-9.8\mathrm{m}/{\mathrm{s}}^2$

Displacement signs for each situation

$$\begin{gathered} {\left(∆y\right)}_a=\mathrm{negative} \\ {\left(∆y\right)}_b=\mathrm{zero} \\ {\left(∆y\right)}_c=\mathrm{positive} \end{gathered}$$

To understand the concept

It is known that horizontal component of velocity is same at any point in projectile motion. So the final speed can be compared by the vertical components of speed. Vertical component can be calculated by using the kinematic equations. In kinematic equations of motion, the motion of an object can be described with constant acceleration. Further it can be ranked according to the speed.

Formulae:

The final velocity can be written as

$v_{fy}^2=v_{0y}^2+\left(2\times a\times ∆y\right)$

To rank situations according to the final speeds of the projectiles just before they land

Rank a;

$$\begin{gathered} v_{fy}^2=v_{0y}^2+\left(2\times -a\times -∆y\right) \\ {v}_{fy}^2=v_{0y}^2+\left(2\times a\times ∆y\right) \end{gathered}$$

So, final speed will be greater than initial speed.

Rank b;

$$\begin{gathered} v_{fy}^2=v_{0y}^2+\left(2\times -a\times 0\right) \\ {v}_{fy}^2=v_{0y}^2 \end{gathered}$$

So, final speed will be same as initial speed.

Rank c;

$$\begin{gathered} v_{fy}^2=v_{0y}^2+\left(2\times -a\times ∆y\right) \\ {v}_{fy}^2=v_{0y}^2-\left(2\times a\times ∆y\right) \end{gathered}$$

So final speed will be less than initial speed.

Therefore, from these three results we can rank the final speed as:

Rank :$a>b>c$