Question

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??

Answer 1

`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`