Question

The position vector of a projectile is given by $\overrightarrow{r}=3t\hat{i}+(4t-5t^2)\hat{j}$. Find the ratio of maximum height and initial speed.
($g=10{\text{m/s}}^2$)

• A. $\frac{4}{25}$
• B. $\frac{4}{15}$
• C. $\frac{8}{25}$
• D. $\frac{8}{15}$

Answer 1

The correct option is A $\frac{4}{25}$

We know that if the projectile is projected at an angle $\theta$ with the horizontal.
$x-$component of displacement is given by
$s_x=u\cos \theta t$
and $y-$component is given by
$s_y=u\sin \theta t-\frac{1}{2}g{t}^2$
So, $\overrightarrow{s}=\left(u\cos \theta \right)t\hat{i}+\left(\right(u\sin \theta )t-\frac{1}{2}g{t}^2)\hat{j}$
Comparing $\overrightarrow{r}$ and $\overrightarrow{s}$
$u\cos \theta =3\text{and}u\sin \theta =4$
$\therefore u^2{\cos }^2\theta +u^2{\sin }^2\theta =3^2+4^2$
$\Rightarrow u^2=25\Rightarrow u=5\text{m/s}$
and $u\cos \theta =3$
$\Rightarrow \cos \theta =\frac{3}{5},\sin \theta =\frac{4}{5}$
Now, $H_{max}=\frac{u^2{\sin }^2\theta }{2g}$
$\therefore$ Required ratio $=\frac{H_{max}}{u}=\frac{u{\sin }^2\theta }{2g}$
$=\frac{5{\left(\frac{4}{5}\right)}^2}{2\times 10}=\frac{16}{5\times 20}=\frac{4}{25}$