/*! 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 93 The projectiles \(A\) and \(B\) ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 93

The projectiles \(A\) and \(B\) thrown with velocities \(v\) and \(\frac{v}{2}\) have the same range. If \(B\) is thrown at an angle of \(15^{\circ}\) to the horizontal, \(A\) must have been thrown at an angle (A) \(\sin ^{-1}\left(\frac{1}{16}\right)\) (B) \(\sin ^{-1}\left(\frac{1}{4}\right)\) (C) \(2 \sin ^{-1}\left(\frac{1}{4}\right)\) (D) \(\frac{1}{2} \sin ^{-1}\left(\frac{1}{8}\right)\)

Short Answer

Expert verified
The angle at which projectile A must have been thrown is \(\sin ^{-1}\left(\frac{1}{4}\right)/2\), hence the correct answer is (D) \(\frac{1}{2} \sin ^{-1}\left(\frac{1}{8}\right)\).

Step by step solution

01

Write down the range formulas for \(A\) and \(B\)

Given that the range is same for \(A\) and \(B\) we have: \(\frac{v^2 sin(2 \Theta_A)}{g} = \frac{((v/2)^2) sin(2 * 15^{\circ})}{g}\). The g on both sides cancels out, leaving: \(v^2 sin(2 \Theta_A) = (v^2/4) sin(30^{\circ})\). This simplifies to: \(4 sin(2 \Theta_A) = sin(30^{\circ})\).
02

Solve for \(2 \Theta_A\)

Rearranging the equation from step 1 gives: \(2 \Theta_A = sin ^{-1}(\frac{1}{4})\).
03

Solve for \(\Theta_A\)

Finally, we can find \(\Theta_A\) by dividing the outcome of step 2 by 2. So we get: \(\Theta_A = sin^{-1}\left(\frac{1}{4}\right)/2\)

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