A projectile is shot at an angle of #pi/12 # and a velocity of #3 2 m/s#. How far away will the projectile land?

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Answer 1 · 2018-04-26T06:12:01

The distance is #=52.25m#

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=32ms^-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*32^2cos^2(1/12pi))=0#

#0.268x-0.0051x^2=0#

#x(0.268-0.0051x)=0#

#x=0#, this is the starting point

#x=0.268/0.0051=52.25m#

graph{0.268x-0.0051x^2 [-5, 68.1, -15.73, 20.8]}