Question:
A particle is projected from the origin in $xy$ plane. Acceleration of particle in $y$ direction is $\alpha$ . Its equation of path of projectile is $y = ax-bx^2$, then initial velocity of the particle is
A particle is projected from the origin in $xy$ plane. Acceleration of particle in $y$ direction is $\alpha$ . Its equation of path of projectile is $y = ax-bx^2$, then initial velocity of the particle is
Updated On: Jul 5, 2022
- $\sqrt{\frac{\alpha}{2b}}$
- $\sqrt{\frac{\alpha(1+a^2)}{2b}}$
- $\sqrt{\frac{\alpha}{a^2}}$
- $\sqrt{\frac{\alpha\,a^2}{b}}$
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The Correct Option is B
Solution and Explanation
y = ax-bx$^2$
$y = x$ tan $\theta - \frac{\alpha \,x^2}{2 u^2 \, \cos^2 \, \theta}$
$\because$ tan $\theta $ = a, $\frac{\alpha}{2u^2 \, \cos^2 \, \theta} = b;$
$\because \, u = \sqrt{\frac{\alpha (1 + a^2)}{2b}}$
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Notes on Motion in a plane
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