A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #5 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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Answer 1 · 2018-07-04T18:48:18

The distance is #=1.349m#

The equation of the trajectory of the projectile in the vertical plane, #O# being the origin is

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

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

The angle of projection is #theta=5/12pi#

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

Therefore,

#y=xtan(5/12pi)-((9.8)/(2*5^2*cos^2(5/12pi)))x^2#

#y=3.732x-2.926x^2#

The highest point is reached when #dy/dx=0#

#dy/dx=3.732-5.852x#

#3.732-5.852x=0#

#=>#, #x=3.732/5.852=0.638#

And

#y=3.732*0.638-2.926*0.638^2=1.189#

The highest point is #=(0.638,1.189)#

The distance from the stat¡rting point is

#d=sqrt(0.638^2+1.189^2)=1.349m#

graph{(3.732x-2.926x^2) [-0.988, 2.43, -0.128, 1.581]}