A projectile is shot at a velocity of #2 m/s# and an angle of #pi/12 #. What is the projectile's maximum height?

1 Answer
Jul 13, 2017

#"maximum height" = 0.0137# #"m"#

Explanation:

We're asked to find the maximum height reached by a projectile given its initial speed and launch angle.

At its maximum height, the object's instantaneous #y#-velocity is #0#. We can use the equation

#(v_y)^2 = (v_(0y))^2 + 2a_yDeltay#

where

  • #v_y# is the #y#-velocity at a certain height #Deltay#. This value is #0# because we're trying to find its maximum height.

  • #v_(0y)# is the initial #y#-velocity.

We find this using the equation

#v_(0y) = v_0sinalpha_0#

where

  • #v_0# is the initial speed (given as #2# #"m/s"#)

  • #alpha_0# is the launch angle, given as #pi/12#

Therefore,

#v_(0y) = (2color(white)(l)"m/s")(sin(pi/12)) = color(red)(0.518# #color(red)("m/s"#

  • #a_y# for any object in free-fall is #-g#, equal to #-9.81# #"m/s"^2#

  • #Deltay# is the height (relative to original height, which is taken to be #0#), which is what we're trying to find

A list of variables we're using:

#v_y = 0#

#v_(0y) = color(red)(0.518# #color(red)("m/s"#

#a_y = -9.81# #"m/s"^2#

#Deltay = ?#

Plugging in known values into the equation, we have

#0 = (color(red)(0.518color(white)(l)"m/s"))^2 - 2(9.81color(white)(l)"m/s"^2)(Deltay)#

#(19.62color(white)(l)"m/s"^2)(Deltay) = 0.268# #"m"^2"/s"^2#

#Deltay = (0.268color(white)(l)"m"^2"/s"^2)/(19.62color(white)(l)"m/s"^2) = color(blue)(0.0137# #color(blue)("m"#

The maximum height reached by this projectile is thus #color(blue)(0.0137# #sfcolor(blue)("meters"#, which makes some sense because it was launched (a) with a small initial speed and (b) with a low angle relative to the horizontal (equivalent to #15^"o"#).