Question

The equation of motion of a projectile is $y=12x-\frac{3}{4}{x}^2$
The horizontal component of velocity is $3m{s}^{-1}$. What is the range of the projectile?

• A. $18m$
• B. $16m$
• C. $12m$
• D. $21.6m$

Answer 1

Answer: (B)

$16m$

Given $y=12x-\frac{3}{4}{x}^2,u_x=3m{s}^{-1}$

$v_y=\frac{d_y}{d_t}=12\frac{dx}{dt}-\frac{3}{2}x\frac{dx}{dt}$

At $x=0,v_y=y_y=12\frac{d2x}{dt}=12u_x=12\times 3=36m{s}^{-1}$

$a_y=\frac{d}{dt}\left(\frac{dy}{dt}\right)$=$12\frac{d^{2}x}{d{t}^2}-\frac{3}{2}$ $[{\left(\frac{dx}{dt}\right)}^2+x\frac{d^{2}x}{d{t}^2}]$

But $\frac{d^{2}x}{d{t}^2}=a_2=0.$,Hence

Range $R=\frac{2u_x{u}_y}{a_y}=\frac{2\times 3\times 36}{27/2}=16m$

Alternatively we have $y=12x-\frac{3}{4}{x}^2$.when projectile again comes to ground $y=0$ and $x=R$.

$0=12R-\frac{3}{4}{R}^2\Rightarrow R=16m$