Beginning with newton's second law of motion, derive the equations of motion for a projectile fired from altitude h above the ground at an angle e to the horizontal and with initial speed equal to vo.

Considering the fire point at (0,h), x-direction positive to the right (→) and y-direction positive to up (↑) and the only force acting after fire is the projectile weight = -mg in the y-direction.

Step-by-step explanation:

First, we apply the Second Newton's Law in both x and y directions:

x-direction:

Integrating we have

Taking into account that a=(dv/dt) and v=(dx/dt):

y-direction:

Integrating we have

Taking into account that a=(dv/dt) and v=(dy/dt):

y=−9/2x+10

step-by-step explanation:

let's solve for y.

0−9x−2y=−20

step 1: add 9x to both sides.

−9x−2y+9x=−20+9x

−2y=9x−20

step 2: divide both sides by -2.

−2y

−2

=

9x−20

−2

y=

−9

2

x+10

sas (side angle side)

aa(angle angle)

sss(side side side)

sas you have to find two sides with an angle in between

aa you find two angles because it will always add up to 180

sss find all sides

step-by-step explanation: