Question

What is the equation of parabolic trajectory of a projectile? ($\theta =$ angle between the projectile motion and the horizontal)

• A. $y=x^{2}tan\theta -\frac{gx}{2u^{2}co{s}^2\theta }$
• B. $y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$
• C. $y=xtan\theta -\frac{g{x}^2}{u^{2}cos2\theta }$
• D. $y=xtan\theta -\frac{g{x}^2}{u^{2}si{n}^2\theta }$

Answer 1

Answer: (B)

$y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$

Let in time $t$, horizontal distance travelled by object is $x$ and vertical distance travelled is $y$.

For horizontal motion:

Since the velocity of object in horizontal direction is constant hence horizontal acceleration $a_x$ will also be zero.

The position of object in horizontal direction at any time $t$ is given by:

$x=x_0+u_{x}t+1/2a_x{t}^2$

where, $x_0=0$ (distance covered at $t=0$)

$a_x=0$

$u_x=u\cos \theta$

Hence, $x=0+u\cos \theta t+1/2\left(0\right)t^2=u\cos \theta$

$t=x/\left(u\cos \theta \right)$ ..................eq(1)

For vertical motion:

Since, the velocity of object in vertical direction is decreasing due to gravity. Hence, vertical acceleration: $a_y=-g$

The position of object in vertical direction at any time $t$ is given by:

$y=y_0+u_{y}t+1/2a_y{t}^2$

where, $y_0=0$ (distance covered at $t=0$)

$a_y=-g$

$u_y=u\sin \theta$

Hence, $y=0+u\sin \theta t+1/2(-g)t^2$

$y=u\sin \theta t-1/2g{t}^2$ .........eq(2)

Putting the value of $t$ from eq1, in eq2, we get:

$y=u\sin \theta \left(\frac{x}{u\cos \theta }\right)-1/2g(\frac{x}{u\cos \theta }{)}^2$

$y=x\tan \theta -\left(\frac{g}{2u^2{\cos }^2\theta }\right)x^2$