Answer:
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.
[tex]\\ x(t)=Vo*cos(e)*t\\ v_x(t)=Vo*cos(e)\\ a_y(t)=0\\ y(t)=h+Vo*sin(e)*t-\frac{g}{2}t^{2}\\ v_x(t)=Vo*sin(e)-gt\\ a_y(t)=-g[/tex]
Step-by-step explanation:
First, we apply the Second Newton's Law in both x and y directions:
x-direction:
[tex]\sum F_x= m\frac{dv_x}{dt} =0[/tex]
Integrating we have
[tex]\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)[/tex]
Taking into account that a=(dv/dt) and v=(dx/dt):
[tex]a_x(t)=\frac{dV_x(t)}{dt}=0\\V_x(t)=\frac{dx(t)}{dt}-->\int\limits^x_0 {} dx = \int\limits^t_0 {Vo*cos(t)} \, dt \\x(t)=Vo*cos(e)*t[/tex]
y-direction:
[tex]\sum F_y= m\frac{dv_x}{dt} =-mg[/tex]
Integrating we have
[tex]\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[/tex]
Taking into account that a=(dv/dt) and v=(dy/dt):
[tex]a_y(t)=\frac{dV_y(t)}{dt}=-g\\V_y(t)=\frac{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-\frac{g}{2}t^{2}[/tex]
