A particle is projected from the ground- If the equation of the trajectory is y - x - x 2-2- then the time of flight isA- 2-gB- 3-g- 9-gD- -2-g

The correct option is A Comparing with standard equation, nbsp;y=x tan θ gx^2/2u^2 cos^2 θ, we get θ=45^0 and u=√(g)/cosθ ⇒ Time of flight,T=2u sin θ/g=2/√(g) ...

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Byju's Answer

Standard XII

Physics

Introduction to Projectile Motion

A particle is...

Question

A particle is projected from the ground. If the equation of the trajectory is y = x - $\frac{x^{2}}{2}$ , then the time of flight is

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Solution

The correct option is A

Comparing with standard equation,

$y = x t a n θ - \frac{g x^{2}}{2 u^{2} c o s^{2} θ}, w e g e t θ = 45^{0} a n d u = \frac{\sqrt{g}}{c o s θ} \Rightarrow T i m e o f f l i g h t, T = \frac{2 u s i n θ}{g} = \frac{2}{\sqrt{g}}$

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A particle is projected from the ground. If the equation of the trajectory is y = x - x22 , then the time of flight is

The correct option is A Comparing with standard equation, y=xtanθ−gx22u2 cos2θ, we get θ=450 and u=√gcosθ ⇒Time of flight,T=2usinθg=2√g

A particle is projected from the ground. If the equation of the trajectory is y = x - $\frac{x^{2}}{2}$ , then the time of flight is

The correct option is A

Comparing with standard equation,

$y = x t a n θ - \frac{g x^{2}}{2 u^{2} c o s^{2} θ}, w e g e t θ = 45^{0} a n d u = \frac{\sqrt{g}}{c o s θ} \Rightarrow T i m e o f f l i g h t, T = \frac{2 u s i n θ}{g} = \frac{2}{\sqrt{g}}$