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Chapter 2: Problem 197

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) \((\mathrm{h} / 2 \mathrm{~g})\)

Short Answer

Expert verified
The short answer is: \(t_1 + t_2 = \frac{2V_{0}}{g}\)

Step by step solution

01

Identify the variables

Let's identify all the known and unknown variables. - Initial velocity (Vo) - Height (h) - Time taken to reach height h (t1 and t2) - Acceleration due to gravity (g)
02

Use the equations of motion

Since the motion is under gravity, we use the equations of motion for an object under gravity. The equation for height (y) in terms of time (t) is given by: \(y = V_{0}t - \frac{1}{2}gt^2\) At point P, the height is h. So, we need to solve for t: \(h = V_{0}t - \frac{1}{2}gt^2\)
03

Solve the equation for t

In order to solve for t, rewrite the equation as a quadratic equation: \(\frac{1}{2}gt^2 - V_{0}t + h = 0\) Now we can use the quadratic formula to solve for t: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In our case, a = 1/2 * g, b = -Vo, and c = h.
04

Calculate the values of t

Plugging these values into the quadratic formula, we get: \(t_1 = \frac{Vo + \sqrt{V_{0}^2 - 2gh}}{g}\) and \(t_2 = \frac{Vo - \sqrt{V_{0}^2 - 2gh}}{g}\) We have to find the sum of these times t1 + t2.
05

Calculate t1 + t2

Add t1 and t2: \(t_1 + t_2 = \frac{Vo + \sqrt{V_{0}^2 - 2gh}}{g} + \frac{Vo - \sqrt{V_{0}^2 - 2gh}}{g}\) Simplify the expression: \(t_1 + t_2 = \frac{2Vo}{g}\)
06

Choose the correct option

Now that we have found the value of t1 + t2, we can match it with the given answer choices. Answer: (B) \(\left[\left(2 \mathrm{v}_{0}\right) / \mathrm{g}\right]\)

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Most popular questions from this chapter

$\mathrm{A}^{\boldsymbol{\longrightarrow}}=2 \mathrm{i} \wedge+2 \mathrm{j} \wedge-\mathrm{k} \wedge\( and \)\mathrm{B}^{\rightarrow}=2 \mathrm{i} \wedge-\mathrm{j} \wedge-2 \mathrm{k} \wedge\( Find \)3 \mathrm{~A}^{\rightarrow}-2 \mathrm{~B}^{\rightarrow}$ (A) \(2 \mathrm{i} \wedge+7 \mathrm{j} \wedge+\mathrm{k} \wedge\) (B) \(2 \mathrm{i} \wedge+8 \mathrm{j} \wedge-\mathrm{k} \wedge\) (C) \(2 \mathrm{i} \wedge+8 \mathrm{j} \wedge+\mathrm{k} \wedge\) (D) \(\mathrm{i} \wedge+7 \mathrm{j} \wedge+\mathrm{k} \wedge\)

A car goes from one end to the other end of a semicircular path of diameter ' \(\mathrm{d}\) '. Find the ratio between path length and displacement. $\begin{array}{lll}\text { (A) }(3 \pi / 2) & \text { (B) } \pi \text { (C) } 2 & \text { (D) } \pi / 2\end{array}$

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) \(8 \mathrm{~ms}^{-1}\)

Comprehensions type questions. A particle is moving in a circle of radius \(R\) with constant speed. The time period of the particle is \(\mathrm{T}\) Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Range of a projectile is \(R\) and maximum height is \(\mathrm{H}\). Find the area covered by the path of the projectile and horizontal line. (A) \((2 / 3) \mathrm{RH}\) (B) \((5 / 3) \mathrm{RH}\) (C) \((3 / 5) \mathrm{RH}\) (D) \((6 / 5) \mathrm{RH}\)

The resultant of two forces of magnitude \(2 \mathrm{~N}\) and \(3 \mathrm{~N}\) can never be. (A) \(4 \mathrm{~N}\) (B) \(1 \mathrm{~N}\) (C) \(2.5 \mathrm{~N}\) (D) \((1 / 2) \mathrm{N}\)
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