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Chapter 2: Problem 194

A particle is to be projected so as to just pass through three equal rings of diameter \(4 \mathrm{~m}\) and placed in parallel vertical plane at distance \(6 \mathrm{~m}\) apart symmetrically with there highest points at a height \(9 \mathrm{~m}\) above the point of projection. If \(\theta\) is angle of projection then find the value of \(\tan \theta\).

Short Answer

Expert verified
The value of \(\tan \theta\) for the given projectile motion is \(\frac{1}{2}\).

Step by step solution

01

Calculate the horizontal distance

The horizontal distance the particle needs to cover is the sum of the distances between the rings. Since there are three rings, there will be two gaps between them, each of distance 6 meters. So, the total horizontal distance (R) can be calculated as: \[R = 2\times 6 = 12\mathrm{~m}\]
02

Calculate the time of flight

The time of flight can be found using the vertical component of the projection. Let's denote the initial upward (vertical) velocity as \(v_0\). We know that: \[v^2 = v_0^2 - 2gh\] We have the highest point at a height of 9 meters, which means the final upward velocity (\(v\)) will be zero at this point. Plugging the values into the equation, we get: \[0 = v_0^2 - 2g(9)\] Solving for \(v_0\): \[v_0 = \sqrt{2g\times 9}\] The time of flight (T) can be calculated from the vertical motion: \[T = \frac{2v_0}{g} = \frac{2\sqrt{2g\times 9}}{g}\]
03

Calculate the horizontal velocity

The horizontal distance covered should be equal to the horizontal velocity multiplied by the time of flight: \[R = V_h \times T\] Plugging in the values of R and T, we can find the horizontal velocity: \[V_h = \frac{12g}{2\sqrt{2g\times 9}}\]
04

Calculate the initial velocity and angle of projection

Now that we have both vertical and horizontal components of the initial velocity, we can find the angle \(\theta\) for the projection using the following relation: \[\tan \theta = \frac{v_0}{V_h}\] Plugging in the values of \(v_0\) and \(V_h\), we get: \[\tan \theta = \frac{\sqrt{2g\times 9}}{\frac{12g}{2\sqrt{2g\times 9}}}\] Simplifying the expression: \[\tan \theta = \frac{1}{2}\] Therefore, the value of \(\tan \theta\) is \(\frac{1}{2}\).

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Most popular questions from this chapter

A particle is thrown with a speed of \(12 \mathrm{~m} / \mathrm{s}\) at an angle \(60^{\circ}\) with the horizontal range. The time interval between the moments when its speed is \(10 \mathrm{~m} / \mathrm{s}\) is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(1.0 \mathrm{~s}\) (B) \(1.2 \mathrm{~s}\) (C) \(1.4 \mathrm{~s}\) (D) \(1.6 \mathrm{~s}\)

A \(2 \mathrm{~m}\) wide truck is moving with a uniform speed \(v_{0}=\) \(8 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed \(v\) when the truck is \(4 \mathrm{~m}\) away from him. The minimum value of \(v\) so that he can cross the road safely is (A) \(2.62 \mathrm{~m} / \mathrm{s}\) (B) \(4.6 \mathrm{~m} / \mathrm{s}\) (C) \(3.57 \mathrm{~m} / \mathrm{s}\) (D) \(1.414 \mathrm{~m} / \mathrm{s}\)

The ceiling of a long hall is \(25 \mathrm{~m}\) high. What is the maximum horizontal distance that a ball thrown with a speed of \(40 \mathrm{~ms}^{-1}\) can go without hitting the ceiling of the hall? (A) \(200 \mathrm{~m}\) (B) \(150 \mathrm{~m}\) (C) \(100 \mathrm{~m}\) (D) \(50 \mathrm{~m}\)

A particle is projected from ground with velocity \(40 \sqrt{2} \mathrm{~m} / \mathrm{s}\) at \(45^{\circ}\). At time \(t=2 \mathrm{~s}\) : (A) Displacement of particle is \(100 \mathrm{~m}\) (B) Vertical component of velocity is \(30 \mathrm{~m} / \mathrm{s}\) (2) with (C) Velocity makes an angle of \(\tan ^{-1}\) horizontal (D) Particle is at height of \(80 \mathrm{~m}\) from ground

For a particle in uniform circular motion, the acceleration \(\vec{a}\) at a point \(\mathrm{P}(R, \theta)\) on the circle of radius \(R\) is (Here \(\theta\) is measured from the \(x\)-axis) (A) \(-\frac{v^{2}}{R} \cos \theta \hat{i}+\frac{v^{2}}{R} \sin \theta \hat{j}\) (B) \(-\frac{v^{2}}{R} \sin \theta \hat{i}+\frac{v^{2}}{R} \cos \theta \hat{j}\) (C) \(-\frac{v^{2}}{R} \cos \theta \hat{i}-\frac{v^{2}}{R} \sin \theta \hat{j}\) (D) \(\frac{v^{2}}{R} \hat{i}+\frac{v^{2}}{R} \hat{j}\)
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