A projectile is shot at an angle of #pi/12 # and a velocity of # 9 m/s#. How far away will the projectile land?

1 Answer
May 24, 2016

Distance to a landing spot is
#V_h*2*V_v/g ~~ 4.1326 (m)#
where
#V_h = 9m/(sec)*cos(pi/12)#
#V_v = 9m/(sec)*sin(pi/12)#

Explanation:

Vector of velocity of a projectile should be represented as a sum of two vectors - horizontal and vertical components.
These components are:
#V_h = 9m/(sec)*cos(pi/12) ~~ 8.69333m/sec#
#V_v = 9m/(sec)*sin(pi/12) ~~2.32937 m/sec#

While the vertical component brings the projectile up, after which it falls down, the horizontal component pulls it forward during the same time.

So, to find the distance to a landing spot, we need the time the projectile is flying.
On its way up the projectile's vertical component of the velocity is diminishing every second by #g m/(sec^2) ~~ 9.8 m/sec^2#.
Therefore, with initial vertical component of #V_v = 9m/(sec)*sin(pi/12)# and a loss of #9.8 m/sec^2# every second, the projectile will continue rising for
#V_v/g ~~ (9sin(pi/12))/9.8 ~~ 0.23769 (sec) #

The same time is needed by the projectile to come back to earth, which brings the total time to go up and down to
#t = 0.23769 * 2 = 0.47538 (sec)#

During the same vertical component was dragging the projectile forward with a speed #V_h#.

So, the distance covered is
#V_h*0.47538=9cos(pi/12)*0.47538 ~~4.1326 m#