Question

A projectile is thrown with a velocity of $50{\text{ms}}^{-1}$ at an angle of ${53}^{\circ }$ with the horizontal. The equation of the trajectory is given
(Take $g=10{\text{m/s}}^2$)

• A. $180y=240x-x^2$
• B. $180y=x^2-240x$
• C. $180y=135x-x^2$
• D. $180y=x^2-135x$

Answer 1

The correct option is A $180y=240x-x^2$

Given that,
Initial velocity of the projectile, $u=50\text{m/s}$
Angle of projectile with horizontal, $\theta ={53}^{\circ }$


The $x-$ component of velocity, $u_x=50\cos {53}^{\circ }=30\text{m/s}$

The $y-$ component of velocity, $u_y=50\sin {53}^{\circ }=40\text{m/s}$

Acceleration in horizontal direction, $a_x=0$

Acceleration in verticle direction$a_y=10{\text{m/s}}^2$(downwards)

Let after time $t$, particle is at point $P(x,y)$
So, $y=u_{y}t-\frac{1}{2}g{t}^2$ ....(1)

and $x=u_x.t\Rightarrow t=\frac{x}{u_x}$ ...(2)

From (1) & (2),

$\Rightarrow y=u_y.\frac{x}{u_x}-\frac{1}{2}\frac{g{x}^2}{u_x^2}$

$\Rightarrow y=\frac{40x}{30}-\frac{1}{2}\times 10\times \frac{x^2}{900}$

$\Rightarrow y=\frac{4x}{3}-\frac{x^2}{180}$

$\Rightarrow 180y=240x-x^2$

Alternate:

Formula for trajectory,
$y=x\tan \theta -\frac{g{x}^2}{2v^2{\cos }^2\theta }$

$y=\frac{4}{3}x-\frac{10\times x^2\times 25}{2\times 50\times 50\times 9}$

$\Rightarrow 180y=240x-x^2$

Hence, option (a) is correct.