Question

The maximum range of projectile is $500m$. If the particle is thrown up a plane, which is inclined at an angle of ${30}^o$ with the same speed, the distance covered by it along the inclined plane will be:

• A. $250m$
• B. $500m$
• C. $750m$
• D. $1000m$

Answer 1

The correct option is B $500m$

Range = $\frac{u^2\sin 2\theta }{g}$ ..(1)

and

$R_{max}=\frac{4^2}{g}$ ... (ii)

From (1) , max rang at $\theta ={45}^{\circ }$

$500=\frac{u^2\times \sin 90}{g}$

$\frac{u^2}{g}=500$

Now,

The distance covered along inclined plane is

$v^2-u^2=2as$

Here $a=-gsin{30}^{\circ }$ and v=0

So,

$0-u^2=2(-gsin{30}^{\circ })s$

$s=\frac{u^2}{g}$

$s=500m$