A projectile is shot at an angle of #pi/12 # and a velocity of #5 m/s#. How far away will the projectile land?

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Answer 1 · 2018-03-31T05:56:05

The range is #=1.28m#

The equation of the trajectory of the projectile in the #(x,y)# plane is

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

The initial velocity is #u=5ms^-1#

The angle is #theta=1/12pi#

The acceleration due to gravity is #=9.8ms^-1#

When the projectile will land when

#y=0#

Therefore,

#xtantheta-(gx^2)/(2u^2cos^2theta)=xtan(1/12pi)-(9.8x^2)/(2*5^2*cos^2(1/12pi))=0#

#x(0.268-0.21x)=0#

#x=0.268/0.21#

#=1.28m#