Answer:
a) [tex]v=19.6 m/s[/tex]
b) [tex]H=19.58 m[/tex]
c) [tex]v_{f}=29.57 m/s[/tex]
Explanation:
a) Let's calculate the work done by the rocket until the thrust ends.
[tex]W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J[/tex]
But we know the work is equal to change of kinetic energy, so:
[tex]W=\Delta K=\frac{1}{2}mv^{2}[/tex]
[tex]v=\sqrt{\frac{2W}{m}}=19.6 m/s[/tex]
b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.
[tex]v_{f}^{2}=v_{i}^{2}-2gh[/tex]
At the maximum height the velocity is 0, so v(f) = 0.
[tex]0=v_{i}^{2}-2gH[/tex]
[tex]H=\frac{19.6^{2}}{2*9.81}=19.58 m[/tex]
c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.
Using the same equation of part b)
[tex]v_{f}^{2}=v_{i}^{2}-2gh[/tex]
[tex]v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s[/tex]
The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.
I hope it helps you!
