Question

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.

Answer 1

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.

`\\ x(t)=Vo*cos(e)*t\\ v_x(t)=Vo*cos(e)\\ a_y(t)=0\\ y(t)=h+Vo*sin(e)*t-(g)/(2)t^(2)\\ v_x(t)=Vo*sin(e)-gt\\ a_y(t)=-g`

Explanation:

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

x-direction:

`\sum F_x= m(dv_x)/(dt) =0`

Integrating we have

`\int\limits^(V_x) _{V_(0x)}{}\, dV_x =\int\limits^(t) _0{0}\, dt\\ V_(0x)=Vo*cos(e)\\ V_x(t)=Vo*cos(e)`

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

`a_x(t)=(dV_x(t))/(dt)=0\\V_x(t)=(dx(t))/(dt)-->\int\limits^x_0 {} dx = \int\limits^t_0 {Vo*cos(t)} \, dt \\x(t)=Vo*cos(e)*t`

y-direction:

`\sum F_y= m(dv_x)/(dt) =-mg`

Integrating we have

`\int\limits^(V_y) _{V_(0y)}{}\, dV_y =\int\limits^(t) _0 {-g} \, dt\\ V_(0y)=Vo*sin(e)\\ V_y(t)=Vo*sin(e)-g*t`

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

`a_y(t)=(dV_y(t))/(dt)=-g\\V_y(t)=(dy(t))/(dt)-->\int\limits^y_h {} dy = \int\limits^t_0 {(Vo*sin(t)-g*t)} \, dt \\y(t)=h+Vo*sin(e)*t-(g)/(2)t^(2)`