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
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
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 freefall 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:
Plugging in known values into the equation, we have
The maximum height reached by this projectile is thus