A projectile is shot at a velocity of # 2 m/s# and an angle of #pi/4 #. What is the projectile's peak height?

1 Answer
Feb 5, 2017

#y_"max"(0.14)~~0.10" meters"# (that's about 4 inches)

Explanation:

You have a projectile and you want to know the vertical position (i.e., the #y#-value) at the projectile's peak. The general equation is

#y=(v_0sin(theta_0))t-1/2 g t^2#

You are given

  • #v_0=2# m/s (pretty slow, about 4.5 mph)
  • #theta_0=pi"/"4#
  • #g=9.8# m/s#""^2#

Plugging in those values we get #y(t)#

#y(t) = (2" m/s")sin(pi"/"4)t-0.5(9.8" m/s"^2)t^2#
#" "=(sqrt(2))t-4.9t^2# meters

A graphing calculator will find your max height at #y~~0.10#
graph{y=2sin(pi/4)x-0.5(9.8)x^2[-.01,.3,-.01,.11]}

The time at which the this function is maximum is discovered by solving for #t# when #dy"/"dt=0#.

#d/dt(y)=y'(t)=sqrt(2)-9.8t#

Set #y'(t)=0#

#sqrt(2)-9.8t=0#

Now solve for #t#
#9.8t=sqrt(2)#
#t=sqrt(2)"/"9.8#
#t~~0.14# seconds

So the maximum height occurs at that time

#y_"max"(0.14)~~(2"m/s")sin(pi"/"4)(0.14"s")-1"/"2(9.8"m/s")(0.14"s")^2#
#y_"max"(0.14)~~0.10" meters"# (that's about 4 inches)