Question

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?

Answer 1

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`