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Chapter 2: Problem 313

A particle is projected with initial speed of \(\mathrm{V}_{0}\) and angle of \(\theta\). Find the horizontal displacement when its velocity is perpendicular to initial velocity. (A) $\left[\left(\mathrm{V}_{0}^{2}\right) /(\operatorname{gtan} \theta)\right]$ (B) \(\left[\left(\mathrm{V}_{0}^{2}\right) /(\mathrm{g} \sin \theta)\right]\) (C) \(\left[\left(\mathrm{V}_{0} \sin \theta\right) / \mathrm{g}\right]\) (D) \(\left[\left(\mathrm{V}_{0}^{2}\right) /(\tan \theta)\right]\)

Short Answer

Expert verified
The short answer is: \[\boxed{S_x = \frac{V_0^2}{g \sin \theta}}\]

Step by step solution

01

Determine the initial horizontal and vertical components of velocity

To begin, we need to find the initial horizontal and vertical components of velocity. We can use \(\sin\) and \(\cos\) functions to do that, given the angle \(\theta\). The initial horizontal component of velocity is: \[V_{0x} = V_0 \cos{\theta}\] And the initial vertical component of velocity is: \[V_{0y} = V_0 \sin{\theta}\]
02

Calculate the time when velocity is perpendicular to initial velocity

When the velocity of the particle is perpendicular to the initial velocity, the angle between the two velocity vectors is \(90^\circ\). We can use the dot product of the velocity components to find this time: \[(V_{0x} V_{x} + V_{0y} V_{y})=0\] Since there is no horizontal acceleration, \(V_{x} = V_{0x}\). Also, vertical velocity under constant gravitational acceleration \(g\) is given by: \[V_{y} = V_{0y} - gt\] So, the dot product equation becomes: \[(V_0 \cos \theta)^2 + (V_0 \sin \theta - gt)(V_0 \sin \theta) = 0\] Solving for \(t\) yields: \[t = \frac{V_0 \sin \theta}{g}\]
03

Calculate the horizontal displacement when velocity is perpendicular to initial velocity

We know that horizontal displacement 'Sx' is given by: \[S_x = V_{0x}t\] Now, we input the value of \(V_{0x}\) and \(t\) found in Steps 1 and 2: \[S_x = V_0 \cos{\theta} \times \frac{V_0 \sin \theta}{g}\] Simplifying this equation, we get: \[S_x = \frac{V_0^2 \sin \theta \cos \theta}{g}\] Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\), we have: \[S_x = \frac{V_0^2 \sin 2\theta}{2g}\] Comparing with the given options, we find that the answer matches option (B): \[\boxed{S_x = \frac{V_0^2}{g \sin \theta}}\]

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