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Chapter 2: Problem 236

A freely falling object travels distance \(\mathrm{H}\). Its velocity is \(\mathrm{V}\). Hence, in travelling further distance of \(4 \mathrm{H}\) its velocity will become ... (A) \(\sqrt{3} \mathrm{~V}\) (B) \(\sqrt{5} \mathrm{~V}\) (C) \(2 \mathrm{~V}\) (D) \(3 \mathrm{~V}\)

Short Answer

Expert verified
The new velocity after falling a total distance of 5H is \(\sqrt{5}V\). Hence, the correct answer is (B) \(\sqrt{5} V\).

Step by step solution

01

Recall the equation of motion

For a freely falling object under the effect of gravity, the equation of motion we can use to relate distance, initial velocity, final velocity, and acceleration is: \(v^2 = u^2 + 2as\), where: - \(v\) is the final velocity, - \(u\) is the initial velocity (which will be 0 since the object starts from rest), - \(a\) is the acceleration due to gravity, and - \(s\) is the displacement. In our case, we will consider two situations: when the object falls a distance of H and a subsequent distance of 4H.
02

Calculate the velocity V after falling distance H

Considering only the first distance, H: \(V^2 = 0^2 + 2aH\). Since we are only interested in the magnitude of the velocities, we can ignore the sign of the acceleration due to gravity: \(V^2 = 2aH\).
03

Calculate the velocity after falling a total distance of 5H

Now, let's call the new final velocity after falling the additional distance of 4H as \(V_{new}\). In this scenario, the object would have fallen a total distance of 5H. So, we have: \(V_{new}^2 = 0^2 + 2a(5H)\). Similary, ignoring the sign of the acceleration due to gravity: \(V_{new}^2 = 10aH\).
04

Find the relation between V and V_{new}

Divide the equation for \(V_{new}^2\) by the equation for \(V^2\): \(\frac{V_{new}^2}{V^2} = \frac{10aH}{2aH} = 5\). Now, take the square root of both sides: \( \frac{V_{new}}{V} = \sqrt{5}\). So, the new velocity after falling a total distance of 5H is \(\sqrt{5}V\). Hence, the correct answer is: (B) \(\sqrt{5} V\).

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Most popular questions from this chapter

Velocity of particle \(\mathrm{A}\) with respect to particle \(\mathrm{B}\) is \(4(\mathrm{~m} / \mathrm{s})\) while they are moving in same direction. And it is \(10(\mathrm{~m} / \mathrm{s})\) while they are in opposite direction. What are the velocities of the particles with respect to the stationary frame of reference. (A) \(7 \mathrm{~ms}^{-1}, 3 \mathrm{~ms}^{-1}\) (B) \(4 \mathrm{~ms}^{-1}, 5 \mathrm{~ms}^{-1}\) (C) \(7 \mathrm{~ms}^{-1}, 4 \mathrm{~ms}^{-1}\) (D) \(10 \mathrm{~ms}^{-1}, 4 \mathrm{~ms}^{-1}\)

The relation between velocity and position of a particle is \(\mathrm{V}=\mathrm{Ax}+\mathrm{B}\) where \(\mathrm{A}\) and \(\mathrm{B}\) are constants. Acceleration of the particle is \(10 \mathrm{~ms}^{-2}\) when its velocity is \(\mathrm{V}\), How much is the acceleration when its velocity is $2 \mathrm{~V}$ (A) \(20 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-1}\) (C) \(5 \mathrm{~ms}^{-2}\) (D) 0

A cartasian equation of a projectile is given by $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$ Calculate its initial velocity. (A) \(\sqrt{10 \mathrm{~ms}^{-1}}\) (B) \(\sqrt{5 \mathrm{~ms}^{-1}}\) (C) \(\sqrt{2} \mathrm{~ms}^{-1}\) (D) \(4 \mathrm{~ms}^{-1}\)

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Which from the following is a scalar? (A) Electric current (B) Velocity (C) acceleration (D) Electric field
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