A projectile is shot from the ground at an angle of #pi/8 # and a speed of #5 /9 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
Aug 1, 2017

#"distance" = 0.0114# #"m"# #= 1.14# #"cm"#

Explanation:

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

Vertical:

When the particle is at its maximum height, its instantaneous #y#-velocity is #0#, and we can use the equation

#(v_y)^2 = (v_0sinalpha_0)^2 - 2gh#

to find the height #h#

Plugging in known values (#v_0 = 5/9# #"m/s"#, #alpha_0 = pi/8#), we have

#0 = ((5/9color(white)(l)"m/s")sin(pi/8))^2 - 2(9.81color(white)(l)"m/s"^2)h#

#h = ul(0.00230color(white)(l)"m"#

The time #t# when this occurs is given by

#v_y = v_0sinalpha_0 - g t#

#0 = (5/9color(white)(l)"m/s")sin(pi/8) - (9.81color(white)(l)"m/s"^2)t#

#t = 0.0217# #"s"#

Horizontal:

We can find the horizontal distance covered by the equation

#x = v_0cosalpha_0t#

We found that #t = 0.0217# #"s"#, so we have

#x = (5/9color(white)(l)"m/s")cos(pi/8)(0.0217color(white)(l)"s")#

#= ul(0.0111color(white)(l)"m"#

Distance:

The distance found via

#r = sqrt(x^2 + h^2) = sqrt((0.0111color(white)(l)"m")^2 + (0.00230color(white)(l)"m")^2)#

#= color(red)(ul(0.0114color(white)(l)"m"# #= color(red)(ul(1.14color(white)(l)"cm"#