91Ó°ÊÓ

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:

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

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