Question #150353
A large gun fires a shell with a muzzle velocity of 400m/sec. What is the renge of the gun when it is elevated at an angle of: 30? and 50?
1
Expert's answer
2020-12-14T07:26:14-0500

Solution:

Range of object projected at an angle θ


R=(V2g)sin(2θ)R=( \dfrac{V^2}{g})sin(2 \theta)


V is initial velocity in ms\tfrac{m}{s}

g is the acceleration of gravity 9.8 (ms2)( \tfrac{m}{s^2})


R1=(40029.8)sin(230)=14140mR_1=( \dfrac{400^2}{9.8} )sin(2*30) = 14140m


R2=(40029.8)sin(250)=16080mR_2=( \dfrac{400^2}{9.8})sin(2*50)=16080m


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