A projectile is shot from the ground at a velocity of #9 m/s# at an angle of #pi/3#. How long will it take for the projectile to land?

1 Answer
Jun 1, 2016

Answer:

It takes #1.59# s.

Explanation:

What makes the projectile to land is the gravity. This is a purely vertical component of the motion.
So we are interested in what the projectile does on the vertical axis.

The velocity vector forms an angle of #pi/3# with the ground, so it has the vertical component given by

#v_v=9*sin(pi/3)=9sqrt(3)/2\approx7.79# m/s.

enter image source here

The equation of motion for an object with free fall is

#s=s_0+v_0t+1/2g t^2#

where #s_0# is the initial position, for us it is zero, #v_0# is the initial velocity, for us it is #v_v#, #g# is the acceleration of gravity #-9.81 m/s^2# and #t# is the time. I use the negative sign for the acceleration because I consider the velocity positive going upward, so the acceleration is negative because it goes downward.

Substituting these values we have

#s=7.79t-1/2 9.81t^2#

We are interested in the position #s=0# because we want to know when the body arrive to the ground (that is the origin of our system of coordinates), so we have

#0=7.79t--4.9051t^2#.

One solution is #t=0# and we expect that at zero time the projectile is still on the ground, then we are interested in the solution with #t\ne0#. Because of this, we can divide both sides for #t# and obtain

#0=7.79-4.9051t#

and the time is then

#t=-7.79/-4.9051\approx1.59# s.