A projectile is shot from the ground at an angle of #pi/4 # 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 · 2016-12-11T09:02:55

The horizontal distance is #=2/g=0.2m#

At the maximum height, the vertical component of the velocity is #=0#

The vertical component of the velocity #=usintheta=2*sin(pi/4)#

#=2*sqrt2/2=sqrt2#

as #cos(pi/4)=sqrt2/2#

The horizontal component of the velocity #=usintheta=2*cos(pi/4)#

#=2*sqrt2/2=sqrt2#

as #sin(pi/4)=sqrt2/2#

We use the equation, #v=u+at#

So, #0=sqrt2-gt#

#t=sqrt2/g#

To find the height, we use #v^2=u^2+2as#

#0=2-2gh#

#h=2/(2g)=1/g#

The horizontal distance is #=u_0*t=sqrt2*sqrt2/g=2/g#

graph{-2.45x^2+x [-0.419, 0.634, -0.063, 0.4635]}