A projectile is shot from the ground at an angle of #pi/12 # and a speed of #2 /3 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

1 Answer
Nov 9, 2017

Split this into two parts, horizontal and vertical

Explanation:

The initial velocity will be split into both horizontal and vertical components. For horizontal, #v_{0x} = v_0Cos(theta)# and for vertical it's #v_{0y} = v_0Sin(theta)#

So, for you #v_{0x}=2/3Cos(pi/12) and v_0y=2/3Sin(pi/12)# Be sure your calculator is in radian mode.

Then, as soon as it launches, gravity will take over, so there will be an acceleration of #-9.8m/s^2#. Keys:

#1 there is no acceleration in the x direction and #v_f = v_0# absent of friction or air resistance.

#2, in the y direction, #v_{fy}=0# because the projectile will come to a momentary stop.

use equations #d=v_0t+1/2at^2 and v_f=v_0+at# as needed for each the x and y components using what you know. The first thing you need to find is the time, noting that the time for x and y is the same because the projectile hits the ground once.

For y: #v_{0y}=2/3m/sSin(pi/12)#
#v_{fy} = 0#
#a_y=-9.8m/s^2#
t=?
d=?

For x: #v_{0x}=2/3m/sCos(pi/12)#
#v_{fx} = v_{0x}#
#a_x=0#
t= the same time for y
d=?