A projectile is shot from the ground at a velocity of #18 m/s# and at an angle of #(3pi)/4#. How long will it take for the projectile to land?

1 Answer
May 29, 2017

Answer:

#2.6"s"#

Explanation:

We're asked to find the time #t# when the projectile hits the ground with a given initial velocity and launch angle. In terms of physics, we need to find the time #t# when the height #y# is zero (there will technically be two times, one time was when the motion first started, and the second time is what we're trying to calculate--when it lands).

To find the time when it lands, we can use the equation

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

#Deltay# will be #0# (the change in its height), and we need to find the initial #y#-velocity, #v_(0y)#. We can find this by

#v_(0y) = v_0sinalpha = 18"m"/"s"sin((3pi)/4) = color(blue)(12.7"m"/"s"#

Plugging in our known values (#g = 9.80"m"/("s"^2)#), we have

#0 = (12.7"m"/"s")t - 1/2(9.80"m"/("s"^2))t^2#

#(4.90"m"/("s"^2))t^2 = (12.7"m"/"s")t#

#(4.90"m"/("s"^2))t = 12.7"m"/"s"#

#color(red)(t = 2.6"s"#

Thus, the projectile will land after #2.6"s"#.