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A projectile is given an initial velocity of \((\hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\), where \(\hat{\mathbf{i}}\) is along the ground and \(\hat{\mathbf{j}}\) is along the vertical. If \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), the equation of its trajectory is [JEE Main 2013] (a) \(y=x-5 x^{2}\) (b) \(y=2 x-5 x^{2}\) (c) \(4 y=2 x-5 x^{2}\) (d) \(4 y=2 x-25 x^{2}\)

Short Answer

Expert verified
(b) The equation of the trajectory is \(y = 2x - 5x^2\).

Step by step solution

01

Identify Initial Velocities

The given initial velocity vector is \(\mathbf{v}_0 = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}\). This means the initial velocity in the horizontal direction \(v_{0x}\) is 1 m/s and in the vertical direction \(v_{0y}\) is 2 m/s.
02

Write Down Equations of Motion

The horizontal motion equation is \(x = v_{0x}t\), which simplifies to \(x = t\) since \(v_{0x} = 1\,\mathrm{m/s}\). The vertical motion is described by \(y = v_{0y}t - \frac{1}{2}gt^2\). Substitute \(v_{0y} = 2\) and \(g = 10\,\mathrm{m/s^2}\) to get \(y = 2t - 5t^2\).
03

Express Time (t) in Terms of x

From the horizontal motion equation, we know \(x = t\). Therefore, the time \(t\) can be expressed as \(t = x\).
04

Substitute Expression of t Into Vertical Motion Equation

Substitute \(t = x\) into the vertical motion equation \(y = 2t - 5t^2\) to express \(y\) in terms of \(x\): \(y = 2x - 5x^2\).
05

Compare With Given Options

The expression obtained, \(y = 2x - 5x^2\), matches option (b). Therefore, the equation of the trajectory of the projectile is \(y = 2x - 5x^2\).

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Key Concepts

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

Equations of Motion
In projectile motion, the path of an object under the influence of gravity can be described using equations of motion. These equations explain how the object's position changes over time.
  • In the horizontal direction, the movement is at a constant velocity, so the equation is simply: \[ x = v_{0x} t \] This tells us that horizontal displacement (\(x\)) grows linearly with time (\(t\)).
  • In the vertical direction, gravity affects the motion, causing the object to accelerate downward. The vertical motion equation is: \[ y = v_{0y} t - \frac{1}{2} g t^2 \] Here, \(y\) represents vertical displacement, \(v_{0y}\) is the initial vertical velocity, and \(g\) is the acceleration due to gravity.
Understanding these equations is crucial because they help us analyze the separate components of a projectile's journey. They enable us to predict where the projectile will be at any given time.
Initial Velocity
The initial velocity of a projectile sets the stage for its entire motion. It is usually given as a vector with two components: horizontal and vertical. In this scenario, the initial velocity is described by \( \mathbf{v}_0 = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} \).
  • The horizontal component, \(v_{0x}\), is 1 m/s. This indicates how fast the projectile moves along the ground.
  • The vertical component, \(v_{0y}\), is 2 m/s. This determines how quickly the projectile rises or falls.
Grasping the concept of initial velocity is essential, as it defines both the speed and direction of a projectile's path from the outset. Identifying these components allows us to separate the motion into horizontal and vertical movements, which can be individually analyzed.
Trajectory Equation
The trajectory equation describes the path that a projectile follows through space. By using the components of the initial velocity and equations of motion, we can derive this equation.In our example:- We know \( x = t \) from the horizontal motion equation.- Substituting \( t = x \) into the vertical equation gives \( y = 2x - 5x^2 \).This shows how the projectile's vertical position \(y\) changes as a function of its horizontal position \(x\). Understanding this trajectory equation is fundamental because it provides a complete picture of the projectile's path. Each term in the equation has a specific role:- The linear term, \(2x\), represents the initial vertical velocity's contribution.- The quadratic term, \(-5x^2\), captures the effect of gravity over time.This form helps predict where the projectile will be at any point along its flight, offering valuable insights for solving related physics problems.

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Most popular questions from this chapter

Two stones are projected with the same velocity in magnitude but making different angles with the horizontal. Their ranges are equal. If the angle of projection of one is \(\frac{\pi}{3}\) and its maximum height is \(y_{1}\), the maximum height of the other will be (a) \(3 y_{1}\) (b) \(2 y_{1}\) (c) \(\frac{y_{1}}{2}\) (d) \(\frac{y_{1}}{3}\)

The ceiling of a long hall is \(25 \mathrm{~m}\) high. Then, the maximum horizontal distance that a ball thrown with a speed of \(40 \mathrm{~m} / \mathrm{s}\) can go without hitting the ceiling of the hall, is [NCERT Exemplar] (a) \(95.5 \mathrm{~m}\) (b) \(105.5 \mathrm{~m}\) (c) \(100 \mathrm{~m}\) (d) \(150.5 \mathrm{~m}\)

Two projectiles thrown from the same point at angles \(60^{\circ}\) and \(30^{\circ}\) with the horizontal attain the same height. The ratio of their initial velocities is (a) 1 (b) 2 (c) \(\sqrt{3}\) (d) \(\frac{1}{\sqrt{3}}\)

A particle is projected with a velocity of \(30 \mathrm{~m} / \mathrm{s}\), at an angle \(\theta_{0}=\tan ^{-1}\left(\frac{3}{4}\right)\). After \(1 \mathrm{~s}\), the particle is moving at an angle \(\theta\) to the horizontal, where \(\tan \theta\) will be equal to \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(]\) (b) 2 (c) \(\frac{1}{2}\) (d) \(\frac{1}{3}\)

A ball is projected from a certain point on the surface of a planet at a certain angle with the horizontal surface. The horizontal and vertical displacement \(x\) and \(y\) vary with time \(t\) in second as \(x=10 \sqrt{3} t\) and \(y=10 t-t^{2}\). The maximum height attained by the ball is (a) \(100 \mathrm{~m}\) (b) \(75 \mathrm{~m}\) (c) \(50 \mathrm{~m}\) (d) \(25 \mathrm{~m}\)

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