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Question: What is the equation of the parabolic trajectory of a projectile? ( \(\theta = \) Angle between the ...

What is the equation of the parabolic trajectory of a projectile? ( θ=\theta = Angle between the projectile motion and the horizontal)
A. y=x2tan⁡θ−gx2u2cos⁡2θy = {x^2}\tan \theta - \dfrac{{gx}}{{2{u^2}{{\cos }^2}\theta }}
B. y=xtan⁡θ−gx22u2cos⁡2θy = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}
C. y=xtan⁡θ−gxu2cos⁡2θy = x\tan \theta - \dfrac{{gx}}{{{u^2}\cos 2\theta }}
D. y=xtan⁡θ−gx2u2sin⁡2θy = x\tan \theta - \dfrac{{g{x^2}}}{{{u^2}{{\sin }^2}\theta }}

Explanation

Solution

In order to solve this question we need to understand the velocity relation with acceleration. So acceleration is defined as a time derivative of velocity or velocity of a body in unit interval of time. And from Newton’s Second law, we get that force is directly proportional to acceleration and it is equal to product of mass and acceleration. So if acceleration is zero then the force in that direction would also be zero, hence velocity in that direction remains constant throughout motion.

Complete step by step answer:
Projectile motion is a motion which is covered by particles thrown in air making some angle with ground.So in this motion, acceleration is only in downward direction and it is equal to acceleration due to gravity,
a⃗y=−gj^{\vec a_y} = - g\hat j
Since there is no acceleration in “x” direction, velocity in that direction remains constant throughout motion. So ax=0{a_x} = 0.Let the initial velocity with which projectile thrown be uu making an angle θ\theta with the horizontal.So initial velocity in x direction is,
ux=ucos⁡θ{u_x} = u\cos \theta
And the initial velocity in y direction is,
uy=usin⁡θ{u_y} = u\sin \theta

Let “t” be the time taken by projectile in moving from ground to some point in air
Using equation of motion in “x” direction we get,
x=uxt+12axt2x = {u_x}t + \dfrac{1}{2}{a_x}{t^2}
Putting values we get,
x=(ucos⁡θ)t→(i)x = (u\cos \theta )t \to (i)
Similarly for motion in “y” direction we have,
y=uyt+12ayt2y = {u_y}t + \dfrac{1}{2}{a_y}{t^2}
Putting values we get,
y=(usin⁡θ)t−12gt2→(ii)y = (u\sin \theta )t - \dfrac{1}{2}g{t^2} \to (ii)
From (i) we get,
t=xucos⁡θt = \dfrac{x}{{u\cos \theta }}
Putting value of “t” in equation (ii) we get,
y=(usin⁡θ)(xucos⁡θ)−12g(xucos⁡θ)2y = (u\sin \theta )(\dfrac{x}{{u\cos \theta }}) - \dfrac{1}{2}g{(\dfrac{x}{{u\cos \theta }})^2}
∴y=xtan⁡θ−gx2u2cos⁡2θ\therefore y = x\tan \theta - \dfrac{{gx}}{{2{u^2}{{\cos }^2}\theta }}

So the correct option is B.

Note: It should be remembered that we have used the equation of motion only due to the fact that acceleration is uniform and constant throughout motion of the projectile. So it is a very useful fact that if we have zero acceleration in some particular direction then the body moves with uniform velocity in that direction.