Answer:
[tex]h=53.09m[/tex] (2)
[tex]v_{min}>5.05m/s[/tex]
[tex]v_{max}<10.4m/s[/tex]
Explanation:
a)Kinematics equation for the first ball:
[tex]v(t)=v_{o}-g*t[/tex]
[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]
[tex]y_{o}=h[/tex] initial position is the building height
[tex]v_{o}=8.9m/s[/tex]
The ball reaches the ground, y=0, at t=t1:
[tex]0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}[/tex]
[tex]h=1/2*g*t_{1}^{2}-v_{o}t_{1}[/tex] (1)
Kinematics equation for the second ball:
[tex]v(t)=v_{o}-g*t[/tex]
[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]
[tex]y_{o}=h[/tex] initial position is the building height
[tex]v_{o}=0[/tex] the ball is dropped
The ball reaches the ground, y=0, at t=t2:
[tex]0=h-1/2*g*t_{2}^{2}[/tex]
[tex]h=1/2*g*t_{2}^{2}[/tex] (2)
the second ball is dropped a time of 1.03s later than the first ball:
t2=t1-1.03 (3)
We solve the equations (1) (2) (3):
[tex]1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}[/tex]
[tex]g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)[/tex]
[tex]g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)[/tex]
[tex]-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)[/tex]
[tex]2.06*gt_{1}-2v_{o}t_{1}=g*1.06[/tex]
[tex]t_{1}=g*1.06/(2.06*g-2v_{o})[/tex]
vo=8.9m/s
[tex]t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s[/tex]
t2=t1-1.03 (3)
t2=3.29sg
[tex]h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m[/tex] (2)
b)[tex]t_{1}=g*1.06/(2.06*g-2v_{o})[/tex]
t1 must : t1>1.03 and t1>0
limit case: t1>1.03:
[tex]1.03>9.81*1.06/(2.06*g-2v_{o})[/tex]
[tex]1.03*(2.06*9.81-2v_{o})<9.81*1.06[/tex]
[tex]20.8-2.06v_{o}<10.4[/tex]
[tex](20.8-10.4)/2.06<v_{o}[/tex]
[tex]v_{min}>5.05m/s[/tex]
limit case: t1>0:
[tex]g*1.06/(2.06*g-2v_{o})>0[/tex]
[tex]2.06*g-2v_{o}>0[/tex]
[tex]v_{o}<1.06*9.81[/tex]
[tex]v_{max}<10.4m/s[/tex]
