A projectile is shot at an angle of #pi/8 # and a velocity of # 43 m/s#. How far away will the projectile land?

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Answer 1 · 2018-04-17T08:46:26

The range is #=133.5m#

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=43ms^-1#

The angle is #theta=(1/8pi)rad#

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

The distance #y=0#

Therefore,

#xtan(1/8pi)-(9.8*x^2)/(2*43^2cos^2(pi/8))=0#

#0.414x-0.0031x^2=0#

#x(0.414-0.0031x)=0#

#x=0#, this is the starting point

#x=0.414/0.0031=133.5m#

graph{0.414x-0.0031x^2 [-0.01, 14.04, -1.347, 5.673]}