Figure 4-26 shows three paths for a football kicked from ground level. Ignoring the effects of air, rank the paths according to (a) time of flight, (b) initial vertical velocity component, (c) initial horizontal velocity component, and (d) initial speed, greatest first.

Trajectory of each path.

$h_1=h_2=h_3$

It deals with the kinematic equations of motion in which the motion can be described with constant acceleration. Also, this problem is based on the concept of projectile motion. When a particle is thrown near the earth’s surface, it travels along a curved path under constant acceleration directed towards the center of the earth surface. We can solve this problem by using kinematic equations of projectile motion.

Formulae:

$$\begin{gathered} v_f^2=v_{0y}^2+\left(2\times a\times ∆y\right) \\ ∆x=v_{0x}\times t+\left(\frac{1}{2}\times a_x\times t^2\right) \end{gathered}$$

Time of flight will be same, because maximum height is same for all.

(Please refer part b for this)

Thus, Rank : 1 = 2 = 3

For vertical motion,

$v_{fy}^2=v_{0y}^2-\left(2gh\right)$

At max height,$v_{fy}=0$

Hence,

$v_{0y}^2=2gh$

As, vertical height is same for all, so initial vertical velocity would be same.

Rank : 1 = 2 = 3

For horizontal motion,$a_x=0$

$∆x=v_{0x}t+\left(\frac{1}{2}{a}_x{t}^2\right)$

Hence,

role="math" localid="1660898925170" $Range=v_{0x}t$

As range $R_3>R_2>R_1$and same time for all, $v_{0x}$will increases with$R$

Therefore,

Rank : 3 > 2 > 1

As initial vertical components of speed are same for all, so initial speed will be in proportion with horizontal component of initial velocity.

Hence,

Rank : 3 > 2 > 1