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?

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Answer 1 · 2016-05-02T21:10:45

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:

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

A projectile is shot from the ground at a velocity of 22 m/s22 m/s and at an angle of (2pi)/3(2pi)/3. How long will it take for the projectile to land?