If a projectile is shot at a velocity of #12 m/s# and an angle of #pi/12#, how far will the projectile travel before landing?

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Answer 1 · 2018-04-07T12:14:32

The range is #=7.34m#

The equation describing the trajectory of the projectile in the #x-y# plane is

#y=xtantheta-(gx^2)/(2u^2cos^2theta)#

The initial velocity is #u=12ms^-1#

The angle is #theta=(1/12pi)rad#

The acceleration due to gravity is #g=9.8ms^-2#

The distance #y=0#

Therefore,

#xtan(1/12pi)-(9.8*x^2)/(2*12^2cos^2(pi/12))=0#

#0.268x-0.0365x^2=0#

#x(0.268-0.0365x)=0#

#x=0#, this is the starting point

#x=0.268/0.0365=7.34m#

graph{0.268x-0.0365x^2 [-0.01, 14.04, -1.347, 5.673]}