Question

The smallest velocity (in $m/s$) with which a particle may be projected in order to have a range of $40m$ on the horizontal plane is
(Take $g=10m/s^2$)

• A. $20$
• B. $30$
• C. $40$
• D. $50$

Answer 1

The correct option is A $20$

Maximum range of projectile from question is $40m$
Hence $R_{max}=40$
Since ramge of projectile, $R=\frac{u^2\sin 2\theta }{g}$
We know at an angle of ${45}^o$ of projection, range of projectile is maximum
Hence, $\frac{u^2}{g}=40$
$u^2=400$
Hence, $u=20m{s}^{-1}$