Answer:
a)[tex]v=6.19m/s[/tex]
b)[tex]v=5.51m/s[/tex]
c)[tex]a=3.3*10^{3}m/s^{2}[/tex]
d)[tex]x=5.78*10^{-3}m[/tex]
Explanation:
h1=195m
h2=1.55
a) Velocity just before the ball strikes the floor:
Conservation of the energy law
[tex]E_{o}=E_{f}[/tex]
[tex]E_{o}=mgh_{1}[/tex]
[tex]E_{f}=1/2*mv^{2}[/tex]
so:
[tex]v=\sqrt{2gh_{1}}=\sqrt{2*9.81*1.95}=6.19m/s[/tex]
b) Velocity just after the ball leaves the floor:
[tex]E_{o}=E_{f}[/tex]
[tex]E_{o}=1/2*mv^{2}[/tex]
[tex]E_{f}=mgh_{2}[/tex]
so:
[tex]v=\sqrt{2gh_{2}}=\sqrt{2*9.81*1.55}=5.51m/s[/tex]
c) Relation between Impulse, I, and momentum, p:
[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ (ma)*t=m(v_{f}-v{o})\\\\ a=\frac{ v_{f}-v{o}}{t}=\frac{ 5.51- (-6.19)}{3.5*10^{-3}}=3.3*10^{3}m/s^{2}[/tex]
d) The compression of the ball:
The time elapsed between the ball touching the ground and it is fully compressed, is half the time the ball is in contact with the ground.
[tex]t_{2}=t/2=3.5/2=1.75ms[/tex]
Kinematics equation:
[tex]x(t)=v_{o}t+1/2*a*t_{2}^{2}[/tex]
Vo is the velocity when the ball strike the floor, we found it at a) 6.19m/s.
a, is the acceleration found at c) but we should to use it with a negative sense, because its direction is negative a Vo, a=-3.3*10^3
So:
[tex]x=6.19*1.75*10^{-3}-1/2*3.3*10^{3}*(1.75*10^{-3})^2=5.78*10^{-3}m[/tex]
