A projectile is shot from the ground at a velocity of #22 m/s# and at an angle of #(2pi)/3#. How long will it take for the projectile to land?

1 Answer
May 2, 2016

The best approach would be to separately look at the y-component of the velocity and treat it as a simple time-of-flight problem.

The vertical component of the velocity is: #22xxcos((2pi)/3-pi/2) " m/s" ~~19.052 " m/s"#

Hence the time of flight for this initial velocity is given as:

#t = (2u)/g = (2xx19.052)/9.8 s ~~3.888 s #