Asked by lesar

an archer standing on a cliff 48m above the level field below shoots an arrow at an angle of 30• above horizontal with a speed of 80m/s.how far from the base of the cliff will the arrow land.
Answered by Henry
Range = Vo^2*sin(2A)/g.
Vo = 80 m/s, A = 30o, g = 9.8 m/s^2.
Answered by Elizabeth
D=V costheta ×t
D=80m/s cos30 ×0.6927
D=48.0m
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