/*! 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 37 A ball is projected from the gro... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball is projected from the ground at angle \(\theta\) with the horizontal. After \(1 \mathrm{~s}\), it is moving at angle \(45^{\circ}\) with the horizontal and after \(2 \mathrm{~s}\) it is moving horizontally. What is the velocity of projection of the ball? (1) \(10 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\) (2) \(20 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\) (3) \(10 \sqrt{5} \mathrm{~ms}^{-1}\) (4) \(20 \sqrt{2} \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The velocity of projection is \(10 \sqrt{5} \mathrm{~m/s}\).

Step by step solution

01

Understand the Problem

A ball is projected at an angle \( \theta \) and after 1 second, it makes a \( 45^\circ \) angle with the horizontal. After 2 seconds, it moves horizontally, meaning its vertical velocity component is zero. We need to find the initial velocity of projection.
02

Analyze Conditions at 2 Seconds

At 2 seconds, the vertical component of velocity \( v_y = 0 \). We know the vertical motion equation \( v_y = u \sin\theta - g \cdot t \). For \( t = 2 \) seconds, \( u \sin\theta - g \cdot 2 = 0 \). Solving for \( u \sin\theta \), we get \( u \sin\theta = 2g \).
03

Analyze Conditions at 1 Second

At 1 second, the angle with the horizontal is \( 45^\circ \), which means \( v_x = v_y \). From this, \( u \cos\theta = u \sin\theta - g \). We already know \( u \sin\theta = 2g \), so substitute into \( u \cos\theta = 2g - g \). So, \( u \cos\theta = g \).
04

Calculate Initial Velocity

Now we have two equations: \( u \sin\theta = 2g \) and \( u \cos\theta = g \). Using these, we find \( u^2 = (u\sin\theta)^2 + (u\cos\theta)^2 = 4g^2 + g^2 = 5g^2 \). Therefore, \( u = \sqrt{5g^2} = g\sqrt{5} \). With \( g = 10 \ \text{m/s}^2 \), \( u = 10\sqrt{5} \ \text{m/s} \).

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Key Concepts

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

Angle of Projection
When discussing projectile motion, the angle of projection is a vital concept. Simply put, this is the angle at which an object is launched or projected above the horizontal plane. It plays a crucial role in determining the trajectory of the object. In our exercise, the angle of projection is denoted by \( \theta \), yet its exact value is not specified. However, significant clues about its effects are given by the behavior of the moving ball: switching angles and eventually moving horizontally.

Mathematically, the angle influences how the initial velocity of the projectile breaks down into vertical and horizontal components. Imagine launching a ball: if you project it straight up, the angle of projection is \( 90^{\circ} \), and if you throw it directly along the ground, it's \( 0^{\circ} \). Angles like \( 45^{\circ} \) are interesting because they often maximize the range, assuming all other factors constant.

In the case of this ball, understanding alterations like moving horizontally after 2 seconds helps decipher the relationship between the angle of projection and resulting trajectory components.
Horizontal and Vertical Components of Velocity
Nothing quite clarifies projectile motion like breaking down velocity into horizontal and vertical components. These components are crucial for predicting the projectile's path over time. Think of the initial velocity as forming the hypotenuse of a right triangle, with these components being the other two sides.

For our problem, initially, the velocity \( u \) can be divided into:
  • Horizontal component: \( u \cos\theta \)
  • Vertical component: \( u \sin\theta \)
With this configuration, the ball’s projection transforms into predictable motion, governed by specific equations.

During the first second, these components help us see the ball at an angle of \( 45^{\circ} \), meaning equal horizontal and vertical velocities. By two seconds, the only component left is horizontal (vertical velocity is zero). So, the component’s initial setup directly affects how the path unfolds. They ensure that any projectile follows the familiar parabolic arc we often observe.
Kinematic Equations
Kinematic equations serve as the backbone of projectile motion analysis. They are a set of formulas that describe motion, helping us calculate various parameters like velocity, time, displacement, and more. In this context, kinematic equations bridge the projection angle \( \theta \), velocity components, and time to map the projectile’s precise movements.

The most pertinent equation here is:\[ v_y = u \sin\theta - g \cdot t\]This formula calculates the vertical velocity \(v_y\) at any time \(t\), factoring in gravity \(g\). When this is set to zero (as in the case of the ball at 2 seconds), key insights into the velocity and angle emerge.

By using kinematic equations for both vertical and horizontal components, we uncover relationships like:
  • At 1 second: \( u \cos\theta = u \sin\theta - g \)
  • At 2 seconds: \( u \sin\theta = 2g \)
Ultimately, these equations lead us to solve for the ball’s initial velocity \( u \), pinpointing the kinetic dynamics of this curved journey.

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Most popular questions from this chapter

A particle has been projected with a speed of \(20 \mathrm{~ms}^{-1}\) at an angle of \(30^{\circ}\) with the horizontal. The time taken when the velocity vector becomes perpendicular to the initial 6 velocity vector is (1) \(4 \mathrm{~s}\) (2) \(2 \mathrm{~s}\) (3) \(3 \mathrm{~s}\) (4) Not possible in this case

Two inclined planes \(O A\) and \(O B\) having inclination (with borizontal) \(30^{\circ}\) and \(60^{\circ}\), respectively, intersect each other at \(O\) as shown in figure. A particle is projected from point \(P\) with velocity \(u=10 \sqrt{3} \mathrm{~ms}^{-1}\) along a direction perpendicular to plane \(O A\). If the particle strikes plane \(O B\) perpendicularly at \(Q\), calculate Time of flight of the particle (1) \(8 \mathrm{~s}\) (2) \(6 \mathrm{~s}\) (3) \(4 \mathrm{~s}\) (4) \(2 \mathrm{~s}\)

Ship \(A\) is located \(4 \mathrm{~km}\) north and \(3 \mathrm{~km}\) east of ship \(B\). Ship \(A\) has a velocity of \(20 \mathrm{~km} \mathrm{~h}^{-1}\) towards the south and ship is moving at \(40 \mathrm{~km} \mathrm{~h}^{-1}\) in a direction \(37^{\circ}\) north of east. Take \(x\)-and \(y\)-axes along east and north directions, respectively. (1) Velocity of \(A\) relative to \(B\) is \(-32 \hat{i}-44 \hat{j}\). (2) Position of \(A\) relative to \(B\) as a function of time is given by \(\vec{r}_{A B}=(3-32 t) \hat{i}+(4-44 t) \hat{j}\) where \(t=0\) when the ships are in position described above. (3) Velocity of \(B\) relative to \(A\) is \(-32 \hat{i}-44 \hat{j}\). (4) At some moment \(A\) will be west of \(B\).

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 truck is moving with a constant velocity of \(54 \mathrm{~km} \mathrm{~h}^{-1}\). In which direction (angle with the direction of motion of truck) should a stone be projected up with a velocity of \(20 \mathrm{~ms}^{-1}\), from the floor of the truck, so as to appear at right angles to the truck, for a person standing on earth? (1) \(\cos ^{-1}\left(-\frac{3}{4}\right)\) (2) \(\cos ^{-1}\left(-\frac{1}{4}\right)\) (3) \(\cos ^{-1}\left(\frac{2}{3}\right)\) (4) \(\cos ^{-1}\left(\frac{3}{4}\right)\)

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