Question

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

• A. $250\, m$.
• B. $500\, m$.
• C. $400\, m$.
• D. $1000\, m$.

Answer 1

Answer: (B)

Since, range is maximum, therefore $\theta=45^{\circ}$.
Hence, using $R=\frac{u^{2} \sin 2 \theta}{g}$
We find $500=\frac{u^{2}}{g}$, i.e. $u=[500 g]^{\frac{1}{2}}$
Distance covered along the inclined plane can be obtained using relation $v^{2}-v_{0}^{2}=2 a x$
$\Rightarrow 0-u^{2}=2 \times\left(-g \sin 30^{\circ}\right) \times x$
$\left[\because a =- g \sin 30^{\circ}\right]$
$\therefore x=\frac{u^{2}}{g}=500\, m$