A projectile is shot from the ground at a velocity of #8 m/s# and at an angle of #(7pi)/12#. How long will it take for the projectile to land?

1 Answer
Jul 21, 2017

Answer:

#t = 1.58# #"s"#

Explanation:

We're asked to find the time #t# of an object's flight, given its initial velocity.

To do this, we recognize that when the object lands (assuming it launches and lands at the same height), its height #Deltay# will be #0#; we can use the equation

#Deltay = v_0sinalpha_0t - 1/2g t^2#

to find this time. For this equation,

  • #Deltay# is the change in height (#0#)

  • #v_0# is the initial speed (#8# #"m/s"#)

  • #alpha_0# is the launch angle (#(7pi)/12#)

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

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

Plugging in known quantities:

#0 = (8color(white)(l)"m/s")sin((7pi)/12)t - 1/2(9.81color(white)(l)"m/s"^2)t^2#

#1/2(9.81color(white)(l)"m/s"^2)t^2 = (7.73color(white)(l)"m/s")t#

#(4.905color(white)(l)"m/s"^2)t = 7.73color(white)(l)"m/s"#

#t = (7.73color(white)(l)"m/s")/(4.905color(white)(l)"m/s"^2) = color(red)(1.58# #color(red)("s"#