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Chapter 2: Problem 311

Equation of a projectile is given by \(\mathrm{y}=\mathrm{Ax}-\mathrm{Bx}^{2}\). Find the range for the particle. (A) \((\mathrm{A} / \mathrm{B})\) (B) \((\mathrm{A} / 4 \mathrm{~B})\) (C) \((\mathrm{A} / 2 \mathrm{~B})\) (D) \((2 \mathrm{~A} / \mathrm{B})\)

Short Answer

Expert verified
The short answer is: Option (A) \(x = \frac{A}{B}\)

Step by step solution

01

Write down the given equation

The equation of the projectile is given as: \(y = Ax - Bx^2\)
02

Set y to zero and solve for x

To find the range of the projectile, we need to find the horizontal distance it covers before hitting the ground. So we set y to 0: \(0 = Ax - Bx^2\) Now let's solve for x: \(Ax = Bx^2\) Divide both sides by x (since x cannot be zero, as the particle would still be at its starting point): \(A = Bx\) Now, let's solve for x: \(x = \frac{A}{B}\)
03

Compare the obtained expression with the given options

We found that x = \(\frac{A}{B}\) Comparing this result with the given options: (A) \( \frac{A}{B} \) (B) \( \frac{A}{4B} \) (C) \( \frac{A}{2B} \) (D) \( \frac{2A}{B} \) From the analysis, we can see that the correct answer is: Option (A) \(x = \frac{A}{B}\)

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Most popular questions from this chapter

Find a unit vector from the followings. (A) \(\hat{\imath}+\hat{j}\) (B) \(\hat{\imath}-\hat{j}\) (C) \((1 / \sqrt{2}) \hat{\imath}+(1 / \sqrt{2}) \hat{\jmath}\) (D) \((1 / \sqrt{2}) \hat{1}-(1 / 2) \hat{\jmath}\)

If resultant of $\mathrm{A}^{\rightarrow}=2 \hat{1}+\hat{\jmath}-\mathrm{k}^{\wedge}, \mathrm{B}^{\rightarrow}=\hat{\imath}-2 \hat{\mathrm{j}}+3 \mathrm{k}^{\text {and }} \mathrm{C}^{\rightarrow}$ is unit vector in y direction, then \(\mathrm{C}^{\rightarrow}\) is (A) \(-\hat{j}\) (B) \(3 \hat{i}-2 \hat{j}+2 \mathrm{k}^{\wedge}\) (C) (D) \(2 \hat{i}+3 \mathrm{k}^{\wedge}\)

Motion of a particle is described by \(\mathrm{x}=(\mathrm{t}-2)^{2}\) Find its velocity when it passes through origin. (A) \(0 \mathrm{~m} / \mathrm{s}\) (B) \(2 \mathrm{~ms}^{-1}\) (C) \(4 \mathrm{~ms}^{-1}\) (D) \(8 \mathrm{~ms}^{-1}\)

Train \(A\) is \(56 \mathrm{~m}\) long and train \(\mathrm{B} 54 \mathrm{~m}\) long. They are travelling in opposite direction with velocity $15(\mathrm{~m} / \mathrm{s})\( and \)5(\mathrm{~m} / \mathrm{s})$ respectively. The time of crossing is. (A) \(12 \mathrm{~s}\) (B) \(6 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(18 \mathrm{~s}\)

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) \(\left[\left(\mathrm{V}_{0}^{2}\right) /(\tan \theta)\right]\)
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