Question

A projectile is shot from the ground at an angle of `(5 pi)/12` and a speed of `6/5 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

`dx~~0.03658m, dy~~0.205m`

Explanation:

I copied, pasted and changed the values of this answer:
https://socratic.org/questions/a-projectile-is-shot-from-the-ground-at-an-angle-of-pi-6-and-a-speed-of-5-m-s-wh#425323
If you have difficulties understanding projectiles motion visit this:
https://socratic.org/physics/2d-motion/projectile-motion

We have to calculate the initial speed in x and in y:
`V_(0x)=V_ocos(theta)=(6/5)cos((5pi)/12)=0.31m/s`
`V_(0y)=V_osin(theta)=(6/5)sin((5pi)/12)=1.159m/s`

Let's calculate the distance in y (height):

We have to know when is the projectile at its maximum height. Let's use the equation of the speed for an object with acceleration:
`V_y=V_(0y)+aDeltat`

The acceleration is `-9.8m/s^2` because of earth's gravity:
`0=1.159+(-9.8)Deltat => Deltat~~0.118s`

To find the distance in y we have to use the equation of distance:
`y=y_(0)+v_(0y)t+1/2 at^2`
`=> y= (1.159)(0.118)+(1/2)(-9.8)(0.118)^2`
`=> y~~0.205m`

To find the distance in x we have to use another time the equation of distance (this time there's no acceleration):
`x=x_(0)+v_(0x)t+1/2 at^2`
`=> x= (0.31)(0.118)`
`=> x~~0.03658m`