Question

A projectile has the maximum range $500m$. 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. $250m$.
• B. $500m$.
• C. $400m$.
• D. $1000m$.

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. $u=[500g{]}^{\frac{1}{2}}$
Distance covered along the inclined plane can be obtained using relation $v^2-v_0^2=2ax$
$\Rightarrow 0-u^2=2\times \left(-g\sin 30^{\circ }\right)\times x$
$\left[\because a=-g\sin 30^{\circ }\right]$
$\therefore x=\frac{u^2}{g}=500m$