If a projectile is shot at an angle of #pi/6# and at a velocity of #23 m/s#, when will it reach its maximum height??

1 Answer
Apr 12, 2017

Answer:

elapsed time to the maximum height is t=1.17 sec.

Explanation:

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The initial velocity #" " v_i" "#can be divided into two part as vertical
and horizontal.

  • The part of #" "v_x" " #supplies the horizontal movement.
  • The part of #" "v_y" "# supplies the vertical movement.

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  • #v_y=v_i*sin theta-g*t#

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  • The #" "v_y" "#part of the #" " v_i# decreases depending on the time.
  • Please notice that the #" "v_y" "# is zero at the maximum height.
  • if we write as #v_y=v_i.sin theta-g*t=0#
  • We get #v_i.sin theta=g.t#
  • if we solve according to t,We get #t=(v_i*sin theta)/g#

  • #t=(23.sin(pi/6))/(9.81)#

  • t=1.17 sec