If a projectile is shot at a velocity of #2 m/s# and an angle of #pi/4#, how far will the projectile travel before landing?

1 Answer
Jan 5, 2016

it will land about 0.638 m far.

Explanation:

We have to write down the equations of motion:

#x= x_0+vcosthetat#
#y= h_0+ v sintheta t + (at^2)/2 #

where #v# is the velocity of the projectile, #theta# is the angle of the trajectory, #h_0# the initial quota of the projectile, #x_0# its initial position.

We have:

  • #h_0=0# and #x_0=0#;
  • #v=2 \ ms^(-1)#
  • #theta= pi/4 #
  • #a=-g \ ~~ -9.81 \ ms^(-2). #

We have #v# and #thetha#: we have to find t from the first equation and substitute its value in the second one.
When the projectile lands, it reaches the ground so we have y=0 at that moment.
The equations of motion at the landing are:

#x= vcosthetat#
#0= v sintheta t - (gt^2)/2 #

from the first equation #t=x/(vcostheta)#.
The other becomes:
# v sin theta * x/(vcostheta) - g/2*(x/(vcostheta))^2=#

Performing the calculation we eventually have:
#x= sqrt( (2v^2 costheta sintheta)/g )~~ 0.638 \ m #