/*! 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 143 A cricketer can throw a ball to ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 143

A cricketer can throw a ball to a maximum horizontal distance of \(100 \mathrm{~m}\). How much high above the ground can the cricketer throw the same ball? (A) \(50 \mathrm{~m}\) (B) \(100 \mathrm{~m}\) (C) \(150 \mathrm{~m}\) (D) \(200 \mathrm{~m}\)

Short Answer

Expert verified
The cricketer can throw the ball to a maximum height of approximately \(25 \mathrm{~m}\) above the ground.

Step by step solution

01

Recall Maximum Horizontal Range Formula

The maximum horizontal range R is achieved when the projectile is launched at an angle of 45 degrees. The formula for maximum horizontal range is given by: \( R = \frac{v^2}{g} \) where v is the initial velocity of the projectile and g is acceleration due to gravity (9.81 m/s虏). First, we need to find the initial velocity of the ball in the horizontal motion.
02

Calculate the initial velocity

According to the information given, the cricketer can throw the ball to a maximum horizontal distance of 100 meters. We can use this to find the initial velocity (v). From the formula above: \( R = \frac{v^2}{g} \) \( 100 = \frac{v^2}{9.81} \) Now, solve for v: \( v^2 = 100 * 9.81 \) \( v = \sqrt{981} \approx 31.3 \mathrm{~m/s} \) The initial velocity of the ball is approximately 31.3 m/s.
03

Calculate the velocity components

Since the ball reaches maximum range at a 45-degree angle, we can find the horizontal and vertical components of the initial velocity. The horizontal component (v_x) and vertical component (v_y) can be given by: \(v_x = v \cdot \cos(45掳) = 31.3 \times \frac{\sqrt{2}}{2} \mathrm{~m/s} \) \(v_y = v \cdot \sin(45掳) = 31.3 \times \frac{\sqrt{2}}{2} \mathrm{~m/s} \)
04

Calculate maximum height

Now we are going to use the vertical component of velocity (v_y) to find the maximum height h. We can use the following kinematic equation: \( v_y^2 = u_y^2 - 2gh \) Here, the final velocity at maximum height is 0 (the upward motion stops momentarily). So, we get: \( 0 = u_y^2 - 2gh \) \( h = \frac{u_y^2}{2g} \) Substitute u_y with v_y, and g with 9.81 m/s虏: \( h = \frac{(31.3\times\frac{\sqrt{2}}{2})^2}{2\times9.81} \) Calculate the value of h: \( h \approx 25 \mathrm{~m} \) So, the cricketer can throw the ball to a maximum height of approximately 25 meters above the ground. Thus, the correct answer is none of the given options.

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 91影视!

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

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

A car is moving on a circular path of radius \(100 \mathrm{~m}\). Its speed \(v\) is changing with time as \(v=2 t^{2}\), where \(v\) in \(\mathrm{ms}^{-1}\) and \(t\) in second. The acceleration of car at \(t=5 \mathrm{~s}\) is approximately (A) \(20 \mathrm{~ms}^{-1}\) (B) \(25 \mathrm{~ms}^{-1}\) (C) \(30 \mathrm{~ms}^{-1}\) (D) \(32 \mathrm{~ms}^{-1}\)

A particle is moving eastwards at a velocity of \(5 \mathrm{~ms}^{-1}\). In \(10 \mathrm{~s}\) the velocity changes to \(5 \mathrm{~ms}^{-1}\) northwards. The average acceleration in this time is (A) \(\frac{1}{2} \mathrm{~ms}^{-2}\) towards north \([2005]\) (B) \(\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}\) towards north-east (C) \(\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}\) towards north-west (D) Zero

A lift is coming from a 8 th floor and is just about to reach the 4 th floor. Taking ground floor as origin and positive direction upward for all quantities, which one of the following is correct? (A) \(x<0, v<0, a>0\) (B) \(x>0, v<0, a<0\) (C) \(x>0, v<0, a>0\) (D) \(x>0, v>0, a<0\)

A boy playing on the roof of a \(10 \mathrm{~m}\) high building throws a ball at a speed of \(10 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) with the horizontal. How far from the throwing point will the ball be at the height of \(10 \mathrm{~m}\) from the ground? \(\left[g=10 \mathrm{~m} / \mathrm{s}^{2}, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]\) \([\mathbf{2 0 0 3}]\) (A) \(5.20 \mathrm{~m}\) (B) \(4.33 \mathrm{~m}\) (C) \(2.60 \mathrm{~m}\) (D) \(8.66 \mathrm{~m}\)
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Solid State Physics

Read Explanation

Electromagnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation

Fields in Physics

Read Explanation

Work Energy and Power

Read Explanation
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.