A particle is projected with speed u at angle - with vertical- Find time of flight-

The correct option is D T= 2ucosθ/g From figure : #160; #160; u_x #160;= u sinθ #160; #160; #160; #160; #160; #160; #16 ...

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Standard XII

Physics

Problem Solving

A particle is...

Question

A particle is projected with speed $u$ at angle $θ$ with vertical. Find time of flight.

T = \frac{u sin θ}{g}

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T = \frac{u cos θ}{g}

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T = \frac{2 u sin θ}{g}

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T = \frac{2 u cos θ}{g}

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Solution

The correct option is D $T = \frac{2 u cos θ}{g}$
From figure : $u_{x} = u s i n θ$ $u_{y} = u c o s θ$
Also $a_{y} = - g$ and $a_{x} = 0$

y direction : $S_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$
As $S_{y} = 0$ for $t = T$ (time of flight)

$∴$ $0 = (u c o s θ) T - \frac{g T^{2}}{2}$ $⟹ T = \frac{2 u c o s θ}{g}$

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A particle is projected with speed u at angle θ with vertical. Find time of flight.

The correct option is D T=2ucosθgFrom figure : ux=usinθ uy=ucosθAlso ay=−g and ax=0y direction : Sy=uyt+12ayt2As Sy=0 for t=T (time of flight)∴ 0=(ucosθ)T−gT22 ⟹T=2ucosθg

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