Which of the path (I) or (II) of a projectile has more time of flight? Use necessary assumptions.
Given $H_{1}$ = $H_{2}$

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Solution

The correct option is C
=

Given $H_{1}$ = $H_{2}$
$\frac{u_{y_{1}}^{2}}{2 g}$ = $\frac{u_{y_{2}}^{2}}{2 g}$ = $\frac{u^{2} s i n^{2} θ_{1}}{2 g}$ $\Rightarrow$ $\frac{u^{2} s i n^{2} θ_{2}}{2 g}$ $\Rightarrow$ $u_{1}$ sin $θ_{1}$ = $u_{2}$ sin $θ_{2}$ or $u_{y_{1}}$ = $u_{y_{2}}$
We know $T_{1} = \frac{2 u_{y_{1}}}{g}$
$T_{2} = \frac{2 u_{y_{2}}}{g}$
Hence both paths take the same time.

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Which of the path (I) or (II) of a projectile has more time of flight? Use necessary assumptions. Given H1 = H2

The correct option is C = Given H1 = H2 u2y12g = u2y22g=u2sin2θ12g ⇒ u2sin2θ22g⇒u1 sin θ1 = u2 sin θ2 or uy1 = uy2 We know T1=2uy1g T2=2uy2g Hence both paths take the same time.

Which of the path (I) or (II) of a projectile has more time of flight? Use necessary assumptions.
Given $H_{1}$ = $H_{2}$