A projectile is shot at an angle of #(5pi)/12 # and a velocity of # 3 m/s#. How far away will the projectile land?

Question

1320 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-30T07:40:23

The range is #=0.46m#

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=3ms^-1#

The angle is #theta=5/12pi#

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(5/12pi)-(9.8x^2)/(2*3^2*cos^2(5/12pi))=0#

#x(3.73-8.13x)=0#

#x=3.73/8.13#

#=0.46m#

graph{3.73x-8.13x^2 [-6.2, 204.7, -42.2, 63.3]}