A projectile is shot at an angle of #pi/8 # and a velocity of # 25 m/s#. How far away will the projectile land?

1 Answer
Apr 14, 2016

Answer:

The projectile will land 45.1 metres away.

Explanation:

Part 1, Work out the half flight time.
In projectile problems we ignore air resistance.  So the trajectory is symmetrical about the maximum height.  If we work out the time it takes to reach maximum height we can double that for the full flight time.

At maximum height the vertical component of the velocity is zero.  I will take upwards as positive, therefore initial velocity will be positive and acceleration will be negative.

Use equation of constant acceleration:
#s_v = ?#
#u_v = 25 sin (π/8)#
#v_v = 0#
#a = -9.8 m.s^(-2)#
#t= t_½ # (I am defining # t_½ # as half of the full flight time.)

Use #v=u + at => t_½ = (v_v - u_v)/a#
#=> t_½ = (0 - 25 sin (π/8))/-9.8 = 0.9762…s#

Part 2, Work out the range of the projectile.
The range is horizontal, so we are considering the horizontal aspect of the motion.  The horizontal velocity is constant for projectiles so we use the constant velocity equation.  But remember that in part 1 we calculated half of the full flight time.  So we must use two multiplied by that value in this equation.
#s_H = u_H × 2t_½ = 25 cos (π/8) × 2 × t_½ = 45.1 m# (3 sf)


Alternative Part 1, Work out the full flight time.
As I said above, projectile trajectories are symmetrical about the maximum height.  So the magnitude of the initial vertical velocity will be equal to that of the final vertical velocity but in the opposite direction, i.e. #u_v = -v_v#.  We can use that to calculate the full flight time directly as follows.

Use equation of constant acceleration:
#s_v = ?#
#u_v = 25 sin (π/8)#
#v_v = -25 sin (π/8)#
#a = -9.8 m.s^(-2)#
#t= t#

Taking up as positive.
Use #v=u + at => t = (v_v - u_v)/a#
#=> t = (-25 sin (π/8) - 25 sin (π/8))/-9.8 = 1.9525…s#

As you can see the value of t is indeed double that for #t_½# which I calculated above.
#t_½ = 0.976 s#
#t = 1.952 s#
#t = 2t_½ #