Question

A body is thrown from a point with speed $50\text{m/s}$ at an angle ${37}^{\circ }$ with horizontal. When it has moved a horizontal distance of $80\text{m}$ then its distance from point of projection is

• A. $40\text{m}$
• B. $40\sqrt{2}\text{m}$
• C. $40\sqrt{5}\text{m}$
• D. None

Answer 1

The correct option is C $40\sqrt{5}\text{m}$

As we know that the equation of trajectory is $y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$, where $y$ is the verticle displacement, $x$ is horizontal displacement.
Putting the values in the equation of trajectory we get,

$y=80\times \frac{3}{4}-\frac{10\times 80\times 80\times 25}{2\times 50\times 50\times 16}=40\text{m}$
$(\tan {37}^{\text{o}}=\frac{3}{4},u=50\text{m/s},x=80\text{m}$ and $\cos {37}^{\text{o}}=\frac{4}{5})$

$\therefore$ Distance from point of projection
$=\sqrt{(80{)}^2+(40{)}^2}\text{m}=40\sqrt{5}\text{m}$