A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #2 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 · 2018-06-23T09:16:52

The distance is #=0.216m#

The initial speed is #u=2ms^-1#

The angle is #theta=5/12pirad#

The equation of the trajectory of the projectile is

#y=xtantheta-(gx^2)/(2u^2cos^2theta)#..............#(1)#

At the maximum height

#dy/dx=0#

#=>#, #dy/dx=tantheta-(2gx)/(2u^2costheta)#

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

Therefore,

#tantheta-(gx)/(u^2cos^2theta)=0#

#x=(u^2tanthetacos^2theta)/(g)=u^2(sinthetacostheta)/g#

and

Plugging this value of #x# in #(1)#

#y=u^2(sinthetacostheta)/g*tantheta-(g/(2u^2cos^2theta))*(u^2(sinthetacostheta)/g)^2#

#=(u^2sin^2theta)/g-1/2(u^2sin^2theta)/g#

#=1/2(u^2sin^2theta)/g#

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

Therefore,

#x=2^2*sin(5/12pi)cos(5/12pi)/9.8=0.102m#

#y=(2^2sin^2(5/12pi))/(2*9.8)=0.190#

The distance from the starting point to the greatest height is

#d=sqrt(x^2+y^2)=0.216m#