Define projectile particle and derive the equation $y\, = \,(\tan \,{\theta _0})x\, - \,\frac{g}{{(2\,\cos \,{\theta _0})}}{x^2}$
Define projectile particle and derive the equation $y\, = \,(\tan \,{\theta _0})x\, - \,\frac{g}{{(2\,\cos \,{\theta _0})}}{x^2}$
When an object is thrown in gravitational field of the Earth, it moves with constant horizontal velocity and constant vertical acceleration. Such a two dimensional motion is called a projectile motion and such an object is called a projectile.
Let the distance travelled by the object at time ' $\mathrm{t}$ ' using with $v_{\mathrm{o}}$ given by $x=\left(v_{0} \cos \theta_{0}\right) t$
Let the distance travelled by the projectile along the y direction be given by
$y=\left(v_{0} \sin \theta_{0}\right) t-1 / 2 g t^{2}$
From $(1)$ $\mathrm{t}=\frac{x}{v_{o} \cos \theta_{o}}$
Putting $\mathrm{t}$ value in $\mathrm{y}$ we get
$y=\left(v_{0} \sin \theta_{0}\right)\left(\frac{x}{v_{o} \cos \theta_{o}}\right)-\frac{1}{2} \mathrm{~g}\left(\frac{x}{v_{o} \cos \theta_{o}}\right)$
$y=x \tan \theta_{o}-\frac{1}{2} \frac{g}{\left(v_{o} \cos \theta_{o}\right)^{2}} \cdot x^{2}$
Similar Questions
Two objects are projected with same velocity ' $u$ ' however at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha+\beta=90^{\circ}$, the ratio of horizontal range of the first object to the $2^{\text {nd }}$ object will be :
Two objects are projected with same velocity ' $u$ ' however at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha+\beta=90^{\circ}$, the ratio of horizontal range of the first object to the $2^{\text {nd }}$ object will be :
- [JEE MAIN 2023]
A ball is thrown upwards and it returns to ground describing a parabolic path. Which of the following remains constant
A ball is thrown upwards and it returns to ground describing a parabolic path. Which of the following remains constant
The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50\,m$. If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be $........\,m$
The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50\,m$. If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be $........\,m$
- [JEE MAIN 2023]
An object is projected with a velocity of $20 m/s$ making an angle of $45^o$ with horizontal. The equation for the trajectory is $h = Ax -Bx^2$ where $h$ is height, $x$ is horizontal distance, $A$ and $B$ are constants. The ratio $A : B$ is ($g = 10 ms^{-2}$)
An object is projected with a velocity of $20 m/s$ making an angle of $45^o$ with horizontal. The equation for the trajectory is $h = Ax -Bx^2$ where $h$ is height, $x$ is horizontal distance, $A$ and $B$ are constants. The ratio $A : B$ is ($g = 10 ms^{-2}$)
A particle is projected from the ground at an angle of $\theta $ with the horizontal with an initial speed of $u$. Time after which velocity vector of the projectile is perpendicular to the initial velocity is
A particle is projected from the ground at an angle of $\theta $ with the horizontal with an initial speed of $u$. Time after which velocity vector of the projectile is perpendicular to the initial velocity is