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 =[latex] \ sqrt{6^{2} + 8^{2}} [/latex]

=[latex] \sqrt{100}[/latex]

=[latex] 10ms^{-1}[/latex]

Angle of projectile [latex]\Theta = tan^{-1}\frac{8}{6} = 53^{\circ}[/latex]

Range of projectile is given by R = [latex] \frac{u^{2}sin2\Theta}{g}[/latex]

R = [latex] \frac{100 * sin106^{\circ}}{10}[/latex]

R = 9.6m

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

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