A projectile is shot from the ground at an angle of #pi/6 # and a speed of #2 m/s#. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

1 Answer
Jul 2, 2017

#"distance" = 0.184# #"m"#

Explanation:

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

When the projectile is at its maximum height, its instantaneous #y#-velocity #v_y# is zero. We can use the equation

#v_y = v_(0y) -g t#

to find the time #t# when this occurs.

First, let's find the components of the initial velocity, for reference:

#v_(0x) = v_0cosalpha_0 = (2color(white)(l)"m/s")cos(pi/6) = color(red)(1.73# #color(red)("m/s"#

#v_(0y) = v_0sinalpha_0 = (2color(white)(l)"m/s")sin(pi/6) = color(green)(1.00# #color(green)("m/s"#

Plugging in our known values into the equation (#v_y = 0# and #g = 9.81# #"m/s"^2#), we have

#0 = color(green)(1.00# #color(green)("m/s") - (9.81color(white)(l)"m/s"^2)t#

#t = 0.102# #"s"#

The #x#- and #y#- positions of at this time are given by

#Deltax = v_(0x)t#

#Deltay = v_(0y)t - 1/2g t^2#

Plugging in known values, we have

#Deltax = (color(red)(1.73)color(white)(l)color(red)("m/s"))(0.102color(white)(l)"s")#

#= 0.177# #"m"#

#Deltay = (color(green)(1.00)color(white)(l)color(green)("m/s"))(0.102color(white)(l)"s") - 1/2(9.81color(white)(l)"m/s"^2)(0.102color(white)(l)"s")^2#

#= 0.0510# #"m"#

The distance from the starting point is thus

#r = sqrt((0.177color(white)(l)"m")^2 + (0.0510color(white)(l)"m")^2)#

#= color(blue)(0.184# #color(blue)("m"#