Question

In a projectile motion, the position vector is given by $\overrightarrow{r}=6t\hat{i}+(8t-5t^2)\hat{j}$; the range is

• A. $10\text{m}$
• B. $10.5\text{m}$
• C. $9.6\text{m}$
• D. $96\text{m}$

Answer 1

The correct option is C $9.6\text{m}$

From the given equation we can write:
$x=6ty=8t-5t^2$

We know the range of the projectile is given by:
$R=\frac{u^2\sin 2\theta }{g}=\frac{2.u\sin \theta .u\cos \theta }{g}$
$R=\frac{2u_x{u}_y}{g}$

As, $u_x=\frac{dx}{dt}=6m/s$
$u_y=\frac{dy}{dt}=8-10t=8m/s$, as $t=0$

$\Rightarrow R=\frac{2\times 6\times 8}{10}=9.6m$