Question

A particle is projected from ground with speed 10 m/s at some angle. If the horizontal range of particle is 10 metres, find minimun speed of particle during the motion. Take $g=10m/s^2$.

• A. 8.4 m/s
• B. 6.8 m/s
• C. 5 m/s
• D. 7.1 m/s

Answer 1

The correct option is D 7.1 m/s

$R=\frac{{\mu }^2}{g}sin2\theta ,u_{min}=ucos\theta \text{(at highest point)}R=\frac{{\mu }^2}{g}sin2\theta 10=\frac{10\times \text{/}10}{\text{/}10}sin2\theta sin2\theta =12\theta ={90}^{\circ }\to \theta ={45}^{\circ }$

$u_{min}=ucos\theta$
$⟹u_{min}=10\cos {45}^{\circ }$
$=10/\sqrt{2}=7.1m/s$