A projectile is shot from the ground at a velocity of #4 m/s# and at an angle of #(5pi)/6#. How long will it take for the projectile to land?

1 Answer
Jul 19, 2017

Answer:

#t = 0.408# #"s"#

Explanation:

We're asked to find the time #t# it takes a projectile to land, given its initial velocity.

When the particle lands, its height will be #0# (assuming it launches and lands at the same height), and we can use the kinematics equation

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

where

  • #Deltay# is the height (#0#)

  • #v_(0y)# is the initial #y#-velocity, equal to

#v_(0y) = v_0sinalpha_0 = (4color(white)(l)"m/s")sin((5pi)/6) = 2# #"m/s"#

  • #t# is the time (what we're trying to find)

  • #g# is the acceleration due to gravity near earth's surface, #9.81# #"m/s"^2#

Plugging in known values, we have

#0 = (2color(white)(l)"m/s")t - 1/2(9.81color(white)(l)"m/s"^2)t^2#

#1/2(9.81color(white)(l)"m/s"^2)t^2 = (2color(white)(l)"m/s")t#

#1/2(9.81color(white)(l)"m/s"^2)t = 2color(white)(l)"m/s"#

#t = (2color(white)(l)"m/s")/(1/2(9.81color(white)(l)"m/s"^2)) = color(red)(0.408# #color(red)("s"#