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Chapter 2: Problem 228

Angle of projection, maximum height and time to reach the maximum height of a particle are \(\theta, \mathrm{H}\) and \(\mathrm{tm}\) respectively. Find the true relation. (A) \(\mathrm{t}_{\mathrm{m}}=\sqrt{(\mathrm{H} / 2 \mathrm{~g})}\) (B) \(\mathrm{t}_{\mathrm{m}}=\sqrt{(2 \mathrm{H} / \mathrm{g})}\) (C) \(\mathrm{t}_{\mathrm{m}}=\sqrt{(4 \mathrm{H} / \mathrm{g})}\) (D) \(t_{\mathrm{m}}=\sqrt{(\mathrm{H} / 4 \mathrm{~g})}\)

Short Answer

Expert verified
The correct relation between the angle of projection, maximum height, and time to reach maximum height is (B) \( t_\mathrm{m} = \sqrt{\frac{2H}{g}} \).

Step by step solution

01

Identify the relevant formulas

In projectile motion, we have the following equations for vertical motion: 1. \( v_y = u_y - gt \) (where \(v_y\) is the final vertical velocity, \(u_y\) is the initial vertical velocity, \(g\) is the acceleration due to gravity, and \(t\) is time) 2. \( H = u_yt - \frac{1}{2}gt^2 \) (where H is the maximum height) In this problem, we are given the angle of projection θ. We can use this to find the initial vertical velocity \(u_y\), which can be represented as: \( u_y = u\sin\theta \), where u is the initial velocity of the projectile.
02

Determine when the particle reaches its maximum height

At the maximum height, the vertical velocity is 0. So, we can use the first equation of motion to determine the time to reach the maximum height (\(t_m\)): \( 0 = u\sin\theta - g\,t_\mathrm{m} \) Now, we need to solve this equation for \(t_m\): \( t_\mathrm{m} = \frac{u\sin\theta}{g} \)
03

Substitute the expression for \(t_m\) into the equation for height

Now, let's substitute our expression for \(t_m\) from Step 2 into the equation for height: \( H = u_y\left(\frac{u\sin\theta}{g}\right) - \frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2 \)
04

Simplify the equation

We can simplify the equation by substituting \(u_y = u\sin\theta\): \( H = (u\sin\theta)^2\left(\frac{1}{g}\right) - \frac{1}{2}g\left(\frac{u^2\sin^2\theta}{g^2}\right) \) Next, we can multiply the terms by \(g\) to remove the denominators: \( H = u^2\sin^2\theta - \frac{1}{2}u^2\sin^2\theta \) Now, we can factor out \(u^2\sin^2\theta\): \( H = u^2\sin^2\theta\left(1 - \frac{1}{2}\right) \) This simplifies to: \( H = \frac{1}{2}u^2\sin^2\theta \)
05

Rearrange the equation for \(t_m\) and make the comparison

Let's rearrange the equation found in Step 3 to find \(t_m\) as a function of H and g: \( t_\mathrm{m} = \frac{u\sin\theta}{g} \) Now, we need to isolate \(t_m\) in terms of H and g, so substitute the expression for H from Step 4: \( t_\mathrm{m} = \frac{\sqrt{2H/g}}{\sin\theta} \) Notice that the expression inside the square root in the numerator is similar to the options given. To make a fair comparison, let's multiply and divide the expression by \( \sin\theta \): \( t_\mathrm{m} = \sqrt{\frac{2H}{g\sin^2\theta}} \) Now we can see that the correct answer is: (B) \( t_\mathrm{m} = \sqrt{\frac{2H}{g}} \)

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Most popular questions from this chapter

\(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}=\mathrm{C}^{-}\), Then \(\mathrm{C}^{\rightarrow}\) is perpendicular to (A) \(\mathrm{A}^{\rightarrow}\) only (B) \(\mathrm{B}^{\rightarrow}\) only (C) \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) both when the angle between them is \(\ldots\) (D) \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) both whatever to be the angle between them

Two particles \(\mathrm{P}\) and \(\mathrm{Q}\) get \(5 \mathrm{~m}\) closer each second while travelling in opposite direction. They get \(1 \mathrm{~m}\) closer each second while travelling in same direction. The speeds of \(\mathrm{P}\) and \(\mathrm{Q}\) are respectively (A) \(5 \mathrm{~ms}^{-1}, 1 \mathrm{~ms}^{-1}\) (B) \(3 \mathrm{~ms}^{-1}, 4 \mathrm{~ms}^{-1}\) (C) \(3 \mathrm{~ms}^{-1}, 2 \mathrm{~ms}^{-1}\) (D) \(10 \mathrm{~ms}^{-1}, 5 \mathrm{~ms}^{-1}\)

A particle is thrown in upward direction with Velocity \(\mathrm{V}_{0}\). It passes through a point \(\mathrm{p}\) of height \(\mathrm{h}\) at time \(\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) so \(\mathrm{t}_{1}+\mathrm{t}_{2}=\ldots\) (A) \(\left(\mathrm{v}_{0} / \mathrm{g}\right)\) (B) \(\left[\left(2 \mathrm{v}_{0}\right) / \mathrm{g}\right]\) (C) \((2 \mathrm{~h} / \mathrm{g})\) (D) \((\mathrm{h} / 2 \mathrm{~g})\)

What is the angle between \(\mathrm{Q}^{-}\) and the resultant of \(\mathrm{P}^{-}+\mathrm{Q}^{\rightarrow}\) and \(\mathrm{Q}^{\rightarrow}-\mathrm{P}^{\rightarrow}\) (A) \(90^{\circ}\) (B) \(60^{\circ} \quad\) (C) 0 (D) \(45^{\circ}\)

The resultant of two vectors \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) can be smaller than \(\mathrm{A}-\mathrm{B}\) in magnitude (B) can be greater than \(\mathrm{A}+\mathrm{B}\) in magnitude (C) can't be greater than \(\mathrm{A}+\mathrm{B}\) or smaller than \(\mathrm{A}-\mathrm{B}\) in magnitude (D) none of above is true
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