Question

The velocity of projection of a projectile is $(6\hat{i}+8\hat{j})m{s}^{-1}$. The horizontal range of the projectile is $(g=10m/s^2)$

• A. $4.9m$
• B. $9.6m$
• C. $19.6m$
• D. $14m$

Answer 1

Answer: (B)

$9.6m$

Velocity of projectile $=\sqrt{6^2+8^2}=\sqrt{100}=10m{s}^{-1}$
Angle of projectile $\theta =ta{n}^{-1}\frac{8}{6}={53}^{\circ }$
Range of projectile is given by $R=\frac{u^{2}sin2\theta }{g}$
$R=\frac{100\times sin{106}^{\circ }}{10}=9.6m$
Hence option $B$ is correct.