Question #299714

Given, Sketch, And Solution of the problem?


Expert's answer

Answer

Range of object projected at an angle θ

R=v2sin2θgR=\frac{v^2\sin2\theta}{g}


Now putting angle of 30° and 50°

So range becomes

R=4002sin2×30°9.81=14140mR=\frac{400^2\sin2\times30°}{9.81}=14140m


And

R=4002sin2×50°9.81=16080mR=\frac{400^2\sin2\times50°}{9.81}=16080m






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