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?

1 Answer
loucololosse
May 18, 2017

dx=0,22 m dy=0,9566

Explanation:

![https://www.ilephysique.net/physique_terminalehttp://-mouvement-projectile-champ-pesanteur.php](https://useruploads.socratic.org/Qg9760t4RO5PXRQNFm0G_physique_terminale-mouvement-projectile-champ-pesanteur_01.gif)

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

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