# The Velocity Of Projectionof A Projectile Is (6Hati) (8Hatj) Ms1. What Is The Horizontal Range Of Of Projectile.(G=10M/S^2)

Velocity of projectile =$\ sqrt{6^{2} + 8^{2}}$

=$\sqrt{100}$

=$10ms^{-1}$

Angle of projectile $\Theta = tan^{-1}\frac{8}{6} = 53^{\circ}$

Range of projectile is given by R = $\frac{u^{2}sin2\Theta}{g}$

R = $\frac{100 * sin106^{\circ}}{10}$

R = 9.6m

Therefore, the horizontal range of a projectile is R = 9.6m

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