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

1 Answer
Jun 3, 2017

Answer:

#2.4# #"s"#

Explanation:

We're asked to find the time when the projectile lands when it is launched with a known initial velocity. We need to find the time #t# when the height #Deltay# is #0#.

We can use the equation

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

to find this. Our known quantities are

  • #Deltay# must be #0#, because we're trying to find the time when this occurs.

  • to find the initial #y#-velocity #v_(0y)#, we can use the equation #v_(0y) = v_ysinalpha#, so #v_(0y) = 12"m"/"s"sin((7pi)/12) = 11.6"m"/"s"#

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

If we make #Deltay# zero, we can rearrange the equation to solve for #t#:

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

#t = (2v_(0y))/g#

Therefore,

#t = (2(11.6cancel("m")/"s"))/(9.8cancel("m")/(cancel("s"^2))) = color(red)(2.4# #color(red)("s"#

The projectile will land after #2.4# seconds.