Question

A particle is projected upwards at an angle $\theta ={45}^{\circ }$ with horizontal from the top of a tower $10\text{m}$ high. Assume the origin is situated at the bottom of the tower and the speed at which the particle is projected is $40\sqrt{2}\text{m/s}$. Find the equation of trajectory of the particle.

• A. $y=x-\frac{x^2}{320}$
• B. $y=x-\frac{x^2}{320}-10$
• C. $y=x-\frac{x^2}{320}+10$
• D. $y=x+\frac{x^2}{320}-10$

Answer 1

The correct option is C $y=x-\frac{x^2}{320}+10$

$u=40\sqrt{2}\text{m/s}$
$\theta ={45}^{\circ }$


For horizontal motion (x-direction),
$u_x=u\cos \theta =40\sqrt{2}\times \cos {45}^{\circ }=40\text{m/s}$
$x$-coordinate of particle, $x$ is given as (initial position $x_i=0$)
$x=u_{x}t$
$\therefore x=40t...\left(1\right)$

For motion in vertical direction (y-direction),
$u_y=u\sin \theta =40\sqrt{2}\times \frac{1}{\sqrt{2}}=40\text{m/s}$
At any time $t$ , y-coordinate of particle is $y$, then displacement in vertical direction($S_y$),
$S_y=y-y_0$ where $y_0=+10\text{m}$
$y_0$ is the initial y-coordinate of particle because origin is at the bottom of the tower and particle is projected from the top of tower.
$\Rightarrow S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
Putting $t=\frac{x}{40}$ from Eq. $\left(1\right)$ and $y_0=10$, $a_y=-g$

$\Rightarrow y-10=(40\times \frac{x}{40})-\frac{1}{2}g{\left(\frac{x}{40}\right)}^2$
$\therefore y=x-\frac{x^2}{320}+10$
Hence, above equation represents the trajectory as it relates the position of particle in $y$ and $x$ direction.