Question

A projectile is shot from the ground at an angle of `pi/6` and a speed of `5 m/s`. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

Answer 1

dx=0,22 m dy=0,9566

Explanation:

We have to calculate the initial speed in x and in y:
`V_(0x)=V_ocos(theta)=(5)cos(pi/6)=sqrt(3)/2`
`V_(0y)=V_osin(theta)=(5)sin(pi/6)=2.5`

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=2.5+(-9.8)Deltat => Deltat=2.5/9.8`

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= (2.5)(2.5/9.8)+(1/2)(-9.8)(2.5/9.8)^2`
`=> y~~0.9566m`

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= (sqrt(3)/2)(2.5/9.8)`
`=> x~~0.22m`