Question

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?

Answer 1

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=?