A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #3/5 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?

1 Answer
Jul 29, 2017

#"distance" = 0.0309# #"m"#

Explanation:

We're asked to find the distance from the launch point a particle is when it reaches its maximum height, given its initial velocity.

To do this, we can use the equations

#(v_y)^2 = (v_0sinalpha_0)^2 - 2g(Deltay)#

and

#Deltax = v_0cosalpha_0t#

to find the vertical and horizontal positions, respectively, when the particle is at its maximum height.

We realize that the instantaneous s#y#-velocity #v_y# is #0# at its maximum height, so we have

#0 = (3/5color(white)(l)"m/s")sin((5pi)/12) - 2(9.81color(white)(l)"m/s"^2)(Deltay)#

#Deltay = color(red)(0.0295# #color(red)("m"#

The time it takes for this to happen is given by

#v_y = v_0sinalpha_0 - g t#

#0 = (3/5color(white)(l)"m/s")sin((5pi)/12) - (9.81color(white)(l)"m/s"^2)t#

#t = 0.0591# #"s"#

We can now use the second equation to determine the horizontal distance #Deltax#:

#Deltax = v_0cosalpha_0t = (3/5color(white)(l)"m/s")cos((5pi)/12)(0.0591color(white)(l)"s")#

#Deltax = color(green)(0.00917# #color(green)("m"#

The distance from the launch point is thus

#r = sqrt((Deltax)^2 + (Deltay)^2) = sqrt((color(green)(0.00917color(white)(l)"m"))^2 + (color(red)(0.0295color(white)(l)"m"))^2)#

#= color(blue)(ul(0.0309color(white)(l)"m"#