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Chapter 5: Problem 78

Jai is standing on the top of a building of height \(25 \mathrm{~m}\) he wants to throw his gun to Veeru who stands on top of another building of height \(20 \mathrm{~m}\) at distance \(15 \mathrm{~m}\) from first building. For which horizontal speed of projection, it is possible? (1) \(5 \mathrm{~ms}^{-1}\) (2) \(10 \mathrm{~ms}^{-1}\) (3) \(15 \mathrm{~ms}^{-1}\) (4) \(20 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
Jai needs a horizontal speed of approximately 15 ms^{-1} to reach Veeru.

Step by step solution

01

Set Up the Problem

Jai needs to throw the gun horizontally to reach Veeru. The gun is initially at a height of 25 meters, and it must travel 15 meters horizontally to reach the second building of height 20 meters.
02

Calculate Time of Fall

Calculate the time it takes for the gun to fall from 25 meters to 20 meters. Use the equation for free fall: \[ h = \frac{1}{2}gt^2 \]where \( h = 5 \ \mathrm{m} \) (since 25m - 20m = 5m) and \( g = 9.8 \ \mathrm{m/s^2} \). Rearrange to find time:\[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 5}{9.8}} \approx 1.01 \ \mathrm{s} \].
03

Calculate Horizontal Speed Requirement

To find the required speed \( v \), use the horizontal distance formula:\[ d = vt \]\( d = 15 \ \mathrm{m} \) and \( t = 1.01 \ \mathrm{s} \) from the previous step. Rearrange to solve for \( v \):\[ v = \frac{d}{t} = \frac{15}{1.01} \approx 14.85 \ \mathrm{ms}^{-1} \].
04

Choose the Appropriate Answer

From the calculation, the closest given answer option to 14.85 is \(15 \ \mathrm{ms}^{-1}\). Therefore, Jai needs a horizontal projection speed of \(15 \ \mathrm{ms}^{-1}\) to reach Veeru.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Projectile Motion
In the realm of physics, when an object is projected horizontally, it enters what's known as horizontal projectile motion. This type of motion occurs when an object moves along a path under the influence of gravity, starting with some initial horizontal velocity. The primary point is that the object is thrown or propelled with no vertical component initially. It follows a curved trajectory due to gravity's effect over time.

For example, in the described exercise, the gun is thrown horizontally by Jai. Important factors include:
  • The gun's initial velocity affects how far it travels before hitting the ground.
  • The only force acting vertically is gravity, causing the gun to fall beneath its initial height over time.
  • Horizontally, it moves with a constant velocity as there's no horizontal acceleration due to gravity.
Free Fall
Free fall describes the motion of an object under the influence of gravitational force only. This means no other forces, like air resistance, are considered.

When we put this into practice for our problem, once the gun leaves Jai's hand, its vertical motion is affected solely by gravity. Initially, the gun is at 25 meters and needs to fall to 20 meters, so the free fall distance is 5 meters. Using the free fall formula:\[ h = \frac{1}{2}gt^2 \]we solve for the time it takes to fall this distance as approximately 1.01 seconds.
  • Free fall under Earth's gravity happens at an acceleration of approximately \(9.8 \ m/s^2\).
  • The time it takes for the object to fall vertically is crucial to determining the required horizontal speed.
Kinematics in Two Dimensions
Kinematics in two dimensions helps us understand motion where an object moves in both horizontal and vertical directions simultaneously. Splitting the trajectory into two components allows us to solve these independently.

For the gun:
  • Horizontal Motion: The gun travels 15 meters horizontally with its velocity constant because there's no acceleration in the horizontal direction.
  • Vertical Motion: The gun experiences free fall due to gravity, affecting how long it stays in the air. The time determined by free fall dictates how long the gun has to reach Veeru horizontally.
By using these components, the required horizontal speed \(v\) is determined through the formula:\[ d = vt \]where \(d\) is the horizontal distance, and \(t\) is the time of flight derived from vertical movement, leading to a calculated speed of 14.85 \( m/s\), closest to 15 \( m/s\) in the options.

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Most popular questions from this chapter

Rain is falling vertically with a velocity of \(25 \mathrm{~ms}^{-1} .\) A woman rides a bicycle with a speed of \(10 \mathrm{~m} \mathrm{~s}^{-1}\) in the north to south direction. What is the direction (angle with vertical) in which she should hold her umbrella to safe herself from rain? (1) \(\tan ^{-1}(0.4)\) (2) \(\tan ^{-1}(1)\) (3) \(\tan ^{-1}(\sqrt{3})\) (4) \(\tan ^{-1}(2.6)\)

A bead is free to slide down on a smooth wire rightly stretched between points \(A\) and \(B\) on a vertical circle of radius \(10 \mathrm{~m}\). Find the time taken by the bead to reach point \(B\), if the bead slides from rest from the highest point \(A\) on the circle.

A river is flowing from west to east at a speed of \(5 \mathrm{mmin}^{-1}\). A man on the south bank of the river, capable of swimming at \(10 \mathrm{~m} \min ^{-1}\) in still water, wants to swim across the river in the shortest time. Finally, he will move in a direction (1) \(\tan ^{-1}(2) \mathrm{E}\) of \(\mathrm{N}\) (2) \(\tan ^{-1}(2) \mathrm{N}\) of \(\mathrm{E}\) (3) \(30^{\circ} \mathrm{E}\) of \(\mathrm{N}\) (4) \(60^{\circ} \mathrm{E}\) of \(\mathrm{N}\)

Points \(A\) and \(C\) are on the horizontal ground and \(A\) and \(B\) are in same vertical plane at a distance of \(1500 \mathrm{~m}\). Simultaneously bullets are fired from \(A, B\) and \(C\) and they collide at \(O\). The bullet at \(B\) is fired at an angle of \(30^{\circ}\) with horizontal towards the ground at velocity \(100 \mathrm{~m} / \mathrm{s}\). The bullet at \(C\) is projected vertically upward at velocity of \(100 \mathrm{~m} / \mathrm{s}\). The bullet projected from \(A\) reaches its maximum height at \(O\). Find the time in which bullets will collide (seconds): (1) 10 (2) 15 (3) 20 (4) 25

A man can swim in still water with a speed of \(2 \mathrm{~ms}^{-1}\). If he wants to cross a river of water current speed \(\sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\) along the shortest possible path, then in which direction should he swim? (1) At an angle \(120^{\circ}\) to the water current (2) At an angle \(150^{\circ}\) to the water current (3) At an angle \(90^{\circ}\) to the water current (4) None of these
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