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Chapter 2: Problem 102

The equation of trajectory of an oblique projectile is \(y=x-\frac{1}{2} x^{2}\). The time of flight of projectile will be (A) \(\frac{2}{\sqrt{g}}\) (B) \(\frac{3}{\sqrt{g}}\) (C) \(\frac{4}{\sqrt{g}}\) (D) \(\frac{2 \sqrt{2}}{\sqrt{g}}\)

Short Answer

Expert verified
The time of flight of the projectile is \(\frac{2}{\sqrt{g}}\), which corresponds to option (A).

Step by step solution

01

Find the Range

Set the equation of trajectory equal to zero and solve for x: \(0 = x - \frac{1}{2}gx^2\), which simplifies to \(2x = gx^2\). Solving for \(x = R\), the range of the projectile, we get \(R = 2 / g\)
02

Determine the Initial Velocity and Angle of Projection

From the equation of trajectory, the coefficient of \(x\) gives us \(u\cos{\theta} = 1\) and the coefficient of \(x^2\) gives us \(u^2\sin{\theta} = \frac{1}{2}g\). From these equations, we infer that the initial velocity \(u = \sqrt{g}\) and the angle of projection \(\theta = 45掳\)
03

Calculate the Time of Flight

Substitute the values for \(R\), \(u\), and \(\cos{\theta}\) into the formula for time of flight: \(T = \frac{2R}{u\cos{\theta}} = \frac{2}{\sqrt{g}}\)

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Most popular questions from this chapter

A particle is projected from origin with speed \(25 \mathrm{~m} / \mathrm{s}\) at angle \(53^{\circ}\) with the horizontal at \(t=0\). Find time of flight.

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Theequationofmotionofa projectile is \(y=12 x-\frac{3}{4} x^{2}\). Given that \(g=10 \mathrm{~ms}^{-2}\), what is the range of the projectile? (A) \(12 \mathrm{~m}\) (B) \(16 \mathrm{~m}\) (C) \(30 \mathrm{~m}\) (D) \(36 \mathrm{~m}\)

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