If a projectile is shot at a velocity of #23 m/s# and an angle of #pi/12#, how far will the projectile travel before landing?

1 Answer
Jun 11, 2017

#27# #"m"#

Explanation:

We're asked to find the horizontal range of the launched projectile with a known initial velocity.

To find this distance, we first must find the time #t# when the object lands, using the equation

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

where

  • #Deltay# is the change in height of the projectile, in #"m"#,

  • #v_(0y)# is the initial #y#-velocity, in #"m"/"s"#. To find this, we must use the equation

#v_(0y) = v_0sinalpha#

where

  • #v_0# is the initial speed (#23"m"/"s"#), and

  • #alpha# is the initial launch angle (#pi/12#)

Therefore, the initial #y#-velocity is

#v_(0y) = (23"m"/"s")sin(pi/12) = 5.95"m"/"s"#

  • #t# is the time, in #"s"#, and

  • #g# is the acceleration due to gravity near Earth's surface, #9.8"m"/("s"^2)#

Plugging in #0# for #Deltay# (because we must find the time when it lands at the same height as it is launched), we have

#0 = (5.95"m"/"s")t - 1/2(9.8"m"/("s"^2))t^2#

#1/2(9.8"m"/("s"^2))t = 5.95"m"/"s"#

#t = color(red)(1.21# #color(red)("s"#

Now, to find how far horizontally it traveled, we use the equation

#Deltax = v_(0x)t#

We must find the initial #x#-velocity now, using the equation

#v_(0x) = v_0cosalpha = 23"m"/"s"cos(pi/12) = 22.2"m"/"s"#

The horizontal range is thus

#Deltax = (22.2"m"/cancel("s"))(color(red)(1.21)cancel(color(red)("s"))) = color(blue)(27# #color(blue)("m"#

rounded to #2# significant figures, the amount given in the problem.