# 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

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

#### Explanation:

**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.

#v_y=v_i*sin theta-g*t#

- 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