A projectile is projected at an angle of 60° to the horizontal with a speed of 60m/s.What will be it's velocity along horizontal.Hence determine the range of the projectile.

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Solution

Horizontal component of velocity=V×cosθ=60×cos60=60×0.5=30m/s

Range=(v^2 × sin2θ)/g
=(60^2 × sin120)/9.8=(3600 × 0.866)/9.8=318.12m

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A projectile is projected at an angle of 60° to the horizontal with a speed of 60m/s.What will be it's velocity along horizontal.Hence determine the range of the projectile.

Horizontal component of velocity=V×cosθ=60×cos60=60×0.5=30m/s Range=(v^2 × sin2θ)/g =(60^2 × sin120)/9.8=(3600 × 0.866)/9.8=318.12m

Horizontal component of velocity=V×cosθ=60×cos60=60×0.5=30m/s

Range=(v^2 × sin2θ)/g
=(60^2 × sin120)/9.8=(3600 × 0.866)/9.8=318.12m