/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 A projectile can have the same r... [FREE SOLUTION] | 91Ó°ÊÓ

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A projectile can have the same range \(R\) for two angles of projection. If \(t_{1}\) and \(t_{2}\) are the times of flight in the two cases, then \(\begin{array}{ll}\text { (A) } t_{1} t_{2} \propto R^{2} & \text { (B) } t_{1} t_{2} \propto \frac{1}{R^{2}}\end{array}\) (C) \(t_{1} t_{2} \propto R\) (D) \(t_{1} t_{2} \propto \frac{1}{R}\)

Short Answer

Expert verified
The product of the times of flight \(t_{1}t_{2}\) is proportional to the range \(R\). So, the correct answer is (C) \(t_{1} t_{2} \propto R\).

Step by step solution

01

Identify Equation for Range of a Projectile

The equation for the range of a projectile, when the projection occurs from a zero level, is \(R = \frac{v^{2} sin(2\Theta)}{g}\). Here, \(v\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(\Theta\) is the angle of projection. Since two angles of projection give the same range, let's take these angles as \(\Theta\) and \((90 - \Theta)\).
02

Calculate Time of Flight

The time of flight for any angle of projection \(\Theta\) is given by \(t = \frac{2v sin(\Theta)}{g}\). Calculate the times of flight \(t_{1}\) and \(t_{2}\) for the angles \(\Theta\) and \((90 - \Theta)\) respectively.
03

Apply Trigonometric Identity

The product of \(t_{1}\) and \(t_{2}\) will result in \(t_{1}t_{2} = \frac{4v^{2}}{g^{2}}(sin\Theta cos\Theta)\). This can be simplified using the trigonometric identity \(2sin\Theta cos\Theta = sin(2\Theta)\) to \(t_{1}t_{2} = \frac{2v^{2}}{g^{2}} sin(2\Theta)\).
04

Relate Time of Flight to Range

By comparing the expression for \(t_{1}t_{2}\) with the formula for the range \(R = \frac{v^2 sin(2\Theta)}{g}\), we can conclude that \(t_{1}t_{2}\) is proportional to \(R\) rather than \(R^{2}\), \(\frac{1}{R^{2}}\) or \(\frac{1}{R}\).

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