If a projectile is shot at a velocity of #2 m/s# and an angle of #pi/6#, how far will the projectile travel before landing?

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Answer 1 · 2018-06-01T13:26:49

The distance is #=0.35m#

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=2ms^-1#

The angle is #theta=(1/6pi)rad#

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

The distance #y=0#

Therefore,

#xtan(1/6pi)-(9.8*x^2)/(2*2^2cos^2(1/6pi))=0#

#0.577x-0.0871x^2=0#

#x(0.577-1.633x)=0#

#x=0#, this is the starting point

#x=0.577/1.633=0.35m#