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Question:
A projectile is given an initial velocity of $(\hat{i} + 2\hat{j}) m/s$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 m/s^2$, then the equation of its trajectory is
A projectile is given an initial velocity of $(\hat{i} + 2\hat{j}) m/s$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 m/s^2$, then the equation of its trajectory is
Updated On: Oct 1, 2024
- $y=x-5x^2$
- $y=2x-5x^2$
- $4y=2x-5x^2$
- $4y=2x-25x^2$
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The Correct Option is B
Solution and Explanation
Initial velocity = (i + 2j)m/s
Magnitude of initial velocity u =$\sqrt{(1)^2+(2)^2}=\sqrt 5 m/s$
Equation of trajectory of projectile is
$ \, \, \, \, y=x tan \theta -\frac{gx^2}{2u^2}(1+tan^2\theta)\bigg[tan \theta=\frac{y}{x}=\frac{2}{1}=2\bigg]$
$\therefore \, \, \, y=x \times 2-\frac{10(x)^2}{2(\sqrt 5)^2}[1+(2)^2]$
$ \, \, \, \, \, \, \, \, \, \, \, \, =2x-\frac{10(x^2)}{2 \times 5}(1+4)$
$ \, \, \, \, \, \, \, \, \, \, \, \, 2x-5x^2$
Magnitude of initial velocity u =$\sqrt{(1)^2+(2)^2}=\sqrt 5 m/s$
Equation of trajectory of projectile is
$ \, \, \, \, y=x tan \theta -\frac{gx^2}{2u^2}(1+tan^2\theta)\bigg[tan \theta=\frac{y}{x}=\frac{2}{1}=2\bigg]$
$\therefore \, \, \, y=x \times 2-\frac{10(x)^2}{2(\sqrt 5)^2}[1+(2)^2]$
$ \, \, \, \, \, \, \, \, \, \, \, \, =2x-\frac{10(x^2)}{2 \times 5}(1+4)$
$ \, \, \, \, \, \, \, \, \, \, \, \, 2x-5x^2$
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration
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