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?

Question

1426 views around the world

You can reuse this answer
Creative Commons License

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#