Question

A projectile has a maximum range of $500m$. If the projectile is now thrown up an inclined plane of ${30}^{\circ }$ with the same velocity, the distance covered by it along the inclined plane will be about

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

Answer 1

The correct option is B $500m$

we know that maximum range of any object projected from an inclined plain is given as,

$R_{max}=\frac{V^2}{g(1+sin\theta )}$

so from here we will find the value of $V$ in meter per second,now the second projectile is projected with same velocity therefore we can again use this velocity for the second case to find the value of range.

so, $R_{max}=\frac{{\left(86.60\right)}^2}{10(1+sin30)}=500meter$