Question

The horizontal and vertical displacements $x$ and $y$ of a projectile at given time $t$ are given by $x=6t\left(m\right)$ and $y=8t-5t^2\left(m\right)$. The range of the projectile in metres is :

• A. $9.6$
• B. $10.6$
• C. $19.2$
• D. $38.2$

Answer 1

The correct option is A $9.6$

Range, $R=u\cos \theta \times t$ (here $x=6tm$)
vertical height, $s=ut+\frac{1}{2}g{t}^2$
Total time is given when displacement, $s=0$
$\Rightarrow 0=u_h\times t-\frac{1}{2}\times 10\times t^2=8t-5t^2$
$\Rightarrow 8t-5t^2=0\Rightarrow t=\frac{8}{5}$
Range $\therefore ,x=6\times \frac{8}{5}=9.6m$