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):

the answer is 844

step-by-step explanation:

it is 844 because if you add 836.50 + 7.5 you will get 844

step-by-step explanation:

what grade level is this? need to know so i can look in my old textbooks!