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
May 31, 2017

Answer:

#0.4# #"s"#

Explanation:

We're asked to find the time #t# when the projectile lands with a known initial velocity and direction. In terms of physics, we need to find the time when the change in #y# (the height) is equal to #0#.

We can use the equation

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

to find the time #t# when #Deltay# is zero,

where

  • #Deltay# is the change in #y#-position (height) of the object, in #"m"#

  • #v_(0y)# is the initial #y#-velocity of the object, in #"m"/"s"#

  • #t# is the time, in #"s"#, and

  • #g# is the acceleration due to gravity near Earth's surface, #9.80"m"/("s"^2)#.

To find the initial #y#-velocity, we can use the equation

#v_(0y) = v_0sinalpha#

where

  • #v_0# is the magnitude of the initial velocity, in #"m"/"s"#, and

  • #alpha# is the initial launch angle.

Plugging in our known variables, we have

#v_(0y) = v_0sinalpha = (4"m"/"s")sin((5pi)/6) = color(red)(2"m"/"s"#

Now that we have our known variables necessary, let's plug them back into the original equation to find the times when the height is #0# (there are two times because the height is zero when it is first launched as well):

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

#0 = (color(red)(2"m"/"s"))t - 1/2(9.80"m"/("s"^2))t^2#

#1/2(9.80"m"/("s"^2))t^2 = (2"m"/"s")t#

We'll eliminate the #t = 0# option by dividing both sides by #t#:

#(4.90"m"/("s"^2))t = (2"m"/"s")#

#t = (2cancel("m")/cancel("s"))/(4.90cancel("m")/("s"^cancel(2))) = color(blue)(0.4"s"#

rounded to one significant figure, which is the amount given in the problem.

Thus, the object will land again ad #t = 0.4"s"#