Given, Sketch, And Solution of the problem?
Answer
Range of object projected at an angle θ
R=v2sin2θgR=\frac{v^2\sin2\theta}{g}R=gv2sin2θ
Now putting angle of 30° and 50°
So range becomes
R=4002sin2×30°9.81=14140mR=\frac{400^2\sin2\times30°}{9.81}=14140mR=9.814002sin2×30°=14140m
And
R=4002sin2×50°9.81=16080mR=\frac{400^2\sin2\times50°}{9.81}=16080mR=9.814002sin2×50°=16080m
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