If a projectile is shot at an angle of #pi/3# and at a velocity of #6 m/s#, when will it reach its maximum height??

1 Answer
Feb 15, 2016

Answer:

#v_f = v_i + at; a=-g; v_f = 0#
#0 = 3sqrt(3) - 9.81t; t=3sqrt(3)/9.81 ~~ .53 s #

Explanation:

  • GIVEN: a projectile shot at angle, #theta = pi/3; v_i = 6m/s #
    find the time require maximum height, i.e. zero velocity at tyhe top.
  • DEFINITION AND PRINCIPLES: here what tou need a kinematic equation that use initial velocity, final velocity and acceleration (gravity in this case)
    #v_(f_y) = v_(i_y) + a_yt; a=-g; v_(f_y)= 0; v_(i_y)=6 sin(pi/3)=3sqrt(3);#
    #0 = 3sqrt(3) - 9.81t; t=3sqrt(3)/9.81 ~~ .53 s #