91Ó°ÊÓ

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

Signs of vertical component of velocity for each situation

$$\begin{gathered} {\mathrm{v}}_{\mathrm{oy}}=70\mathrm{upward} \\ {\mathrm{v}}_{\mathrm{oy}}=70\mathrm{uproad} \\ {\mathrm{v}}_{\mathrm{oy}}=70\mathrm{download} \\ {\mathrm{v}}_{\mathrm{oy}}=70\mathrm{download} \end{gathered}$$

The problem deals with the magnitude of the vector. Also it involves the kinematic equation of motion in which the motion of an object can be described with constant acceleration. Magnitude of vectors can be calculated by using components of each vector. Time for each vector can be calculated by using kinematic equation and then it can be compared.

Formulae:

The displacement in kinematic equation can be written as,

$∆y=\left(V_{0y}×t\right)+\left(\frac{1}{2}×a×t^2\right)$

Components of each vector have same magnitude, so launch speed would be same for each vector.

Therefore, the rank of vectors according to the launch speed of the projectile

Rank:1=2=3=4

Now we will calculate time of flight for each vector by using kinematic equation for vertical motion.

$∆y=\left(V_{0y}×t\right)+\left(\frac{1}{2}×a×t^2\right)$

For vector 1 and 2

Note: consider downward direction as positive for calculations.

Vertical component of initial velocity is upward.

$∆y=\left(-V_{0y}×t\right)+\left(\frac{1}{2}×9.8×t^2\right)$

$4.9t^2-V_{0y}-∆y=0$ ……. (i)

For vector 3 and 4:

Vertical component of initial velocity is downward.

$∆y=\left(-V_{0y}×t\right)+\left(\frac{1}{2}×9.8×t^2\right)$

$4.9t^2-V_{0y}-∆y=0$…… (ii)

From equation (i) and (ii) we can say that, time of flight from equation (i) is greater than from equation (ii).

Therefore, the rank of vectors according to the time of flight of the projectile is:

Rank 1=2>3=4