Question

A particle of mass $1\text{kg}$ is projected with a speed of $10\text{m/s}$ in a uniform gravitational field as shown in the figure. The trajectory of the particle is given by

• A. $9x=24y+y^2$
• B. $18x=24y+5y^2$
• C. $18x=12y+10y^2$
• D. $36x=24y-5y^2$

Answer 1

The correct option is B $18x=24y+5y^2$

Let us, break the velocity into horizontal and vertical components.


Velocity component along $x-$axis,
$u_x=10\cos {37}^{\circ }=8\text{m/s}$

Velocity component along $y-$axis,
$u_y=10\sin {37}^{\circ }=6\text{m/s}$

Accelerations along the $x-$axis and $y-$ can be calculated by,

$a_x=\frac{F}{m}=E_x=20\text{N/kg}$

$a_y=E_y=0$

Let, the coordinates of the particle after time $t$ be $(x,y)$,

Applying the equation of motion along $x-$axis, we have

$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$\Rightarrow x=8t+\frac{1}{2}\times 20\times t^2$

$\Rightarrow x=8t+10t^2...\left(1\right)$

Applying the equation of motion along $y-$axis, we have

$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$\Rightarrow y=6t+\frac{1}{2}\times 0\times t^2$

$\Rightarrow t=\frac{y}{6}...\left(2\right)$

using (1) and (2) we get,

$x=8\times \frac{y}{6}+10{\left(\frac{y}{6}\right)}^2$

$\Rightarrow x=\frac{4y}{3}+\frac{5y^2}{18}$

$\Rightarrow 18x=24y+5y^2$

Hence, option (b) is the correct answer.

Why this question: This question requires the analytical skills. It helps the students to correlate the concept of motion in 2D and gravitation.