Answer:
[tex]v(7s)=46.38 m/s[/tex]
[tex]t(0m)=69.88s [/tex] (impact)
[tex]v(0m)=-58 m/s[/tex] (impact)
Explanation:
The equation for height is [tex]s(t)=58t-0.83t^2[/tex]
We derive the equations for velocity and acceleration by derivating it respect with time:
[tex]v(t)=\frac{ds(t)}{dt}=58-2(0.83)t=58-1.66t[/tex]
[tex]a(t)=\frac{dv(t)}{dt}=-1.66[/tex]
So we have (adding units at the end):
The velocity after 7s (evaluating the velocity formula for 7):
[tex]v(7)=58-1.66(7)[/tex]
[tex]v=46.38 m/s[/tex]
The time it hits the moon (when the height is null again):
[tex]s(t)=0=58t-0.83t^2[/tex]
One solution is t=0, which corresponds to departure of course, but we want the other solution, which is for impact.
[tex]0=58-0.83t[/tex]
[tex]0.83t=58[/tex]
[tex]t=69.88s [/tex]
The velocity at the time it hits the moon:
[tex]v(69.88)=58-1.66(69.88)[/tex]
[tex]v=-58 m/s[/tex]
As it should be since the arrow left the moon at 58m/s.
