/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 A stone is allowed to fall from ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A stone is allowed to fall from the top of a tower and cover half the height of the tower in the last second of its journey. The time taken by the stone to reach the foot of the tower is (A) \((2-\sqrt{2}) \mathrm{s}\) (B) \(4 \mathrm{~s}\) (C) \((2+2 \sqrt{2}) \mathrm{s}\) (D) \((2+\sqrt{2}) \mathrm{s}\)

Short Answer

Expert verified
The total time taken by the stone to reach the foot of the tower is \((2 + \sqrt{2})s\).

Step by step solution

01

Write down the given information

We are given that the stone falls from a tower of height H, and it covers half the height (H/2) in the last second. The initial velocity is 0 m/s as the stone is at rest, and the acceleration due to gravity is 9.8 m/s².
02

Calculate time for the last second

Using the first equation, as described in the analysis: \[h = ut + \frac{1}{2}at^2\] We can plug in the values for h, u, and a for the last second to find out the time taken: \[\frac{H}{2} = 0 \times (1) + \frac{1}{2}(9.8)(1)^2\] This simplifies to: \[\frac{H}{2} = 4.9\]
03

Calculate time for the first part of the fall

Now we need to find the time taken for the stone to fall to the halfway point of the tower. Let this time be t. We have the following equation: \[H - \frac{H}{2} = ut + \frac{1}{2}at^2\] The equation becomes: \[\frac{H}{2} = 0 \times t + \frac{1}{2}(9.8)t^2\] Plug in the value of H/2, which is 4.9: \[4.9 = \frac{1}{2}(9.8)t^2\]
04

Solve for t

To solve for t, divide both sides of the equation by 4.9: \[1 = \frac{1}{2}(9.8)t^2\] Now, multiply both sides of the equation by 2 to get: \[2 = (9.8)t^2\] Divide by 9.8: \[\frac{2}{9.8} = t^2\] Then, take the square root of both sides to get the value of t: \[t = \sqrt{\frac{2}{9.8}}\]
05

Calculate the total time

The total time will be the sum of the time it takes to fall to the halfway point and the last second: Total time = t (time for first part) + 1 (last second) Total time = \(\sqrt{\frac{2}{9.8}} + 1\)
06

Simplify and match the answer with the given options

To match the answer with the given options, find the common denominator for both terms and obtain: Total time = \(\frac{\sqrt{2}}{\sqrt{9.8}} + \frac{\sqrt{9.8}}{\sqrt{9.8}}\) Total time = \(\frac{\sqrt{2}+\sqrt{9.8}}{\sqrt{9.8}}\) Now, we can see that our answer matches option (D): Total time = \((2 + \sqrt{2})s\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two boys (at ground) simultaneously aim their guns at a bird sitting on a tower. The first boy releases his shot with speed of \(100 \sqrt{2} \mathrm{~m} / \mathrm{s}\) at an angle \(45^{\circ}\) with the horizontal. The second boy is behind the first boy by a distance \(\quad 100(\sqrt{3}-1) \mathrm{m}\) and releases his shot with speed \(200 \mathrm{~m} / \mathrm{s}\). Both the shots hit the bird simultaneously. Horizontal distance of the foot of tower from first boy is (A) \(50 \mathrm{~m}\) (B) \(75 \mathrm{~m}\) (C) \(100 \mathrm{~m}\) (D) None of these

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) \(\frac{v^{2}}{R} \hat{i}+\frac{v^{2}}{R} \hat{j}\)

The position \((x)\) of a particle varies with time as \(x=\) \(t^{3}-3 t^{2}\), where \(t\) is time in second. Match Column-I with Column-II. $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \text { (A) Position of particle is } & \text { (1) } t=3 \mathrm{~s} \\\ \text { zero at } \\ \text { (B) Velocity of particle is } & \text { (2) } t=2 \mathrm{~s} \\ \text { zero at } \\ \text { (C) Magnitude of } & \text { (3) } t=5 \mathrm{~s} \\ \text { acceleration of particle } \\ \text { is zero at } \\ \begin{array}{ll} \text { (D) Magnitude of } \\ \text { instantaneous } \\ \text { acceleration is equal to } \\ 18 \mathrm{~m} / \mathrm{s}^{2} \end{array} & \text { (4) } t=1 \mathrm{~s} \\ \hline \end{array} $$

A stone is projected from the ground with velocity \(50 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ} .\) It crosses a wall after \(3 \mathrm{~s}\). How far beyond the wall the stone will strike the ground \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(90.2 \mathrm{~m}\) (B) \(89.6 \mathrm{~m}\) (C) \(86.6 \mathrm{~m}\) (D) \(70.2 \mathrm{~m}\)

The co-ordinates of a moving particle at any time \(t\) are given by \(x=\alpha t^{3}\) and \(y=\beta t^{3}\). The speed of the particle at time \(t\) is given by (A) \(3 t \sqrt{\alpha^{2}+\beta^{2}}\) (B) \(3 t^{2} \sqrt{\alpha^{2}+\beta^{2}}\) (C) \(t^{2} \sqrt{\alpha^{2}+\beta^{2}}\) (D) \(\sqrt{\alpha^{2}+\beta^{2}}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.