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?

Question

1054 views around the world

You can reuse this answer
Creative Commons License

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]}