Question

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??

Answer 1

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