A soccer ball is dropped from a hot air balloon at a height of 1000 ft. Two seconds after the first soccer ball is dropped, a tennis ball is projected vertically upward from the ground with a velocity of 248 ft/sec. When and where will the stones pass each other?
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Expert's answer
2020-10-09T07:00:38-0400
Given:
t=2 s
H=1000 ft
v=248 ft/sec
Calculate at what height the soccer ball will be after t=2 seconds, this give us our reference height:
h=2gt2.
At the height of the first ball
y=H−h=H−2gt2
the tennis ball is launched and goes upward.
At some height above the ground x after time T the objects will meet. By the time T, the soccer ball will make
H−x=uT+2gT2==T2gh+2gT2==g(tT+2T2).
The tennis ball will be at the same height at the same time, but its motion will be defined by another equation:
x=vT−2gT2.
Substitute for x in the above equation:
H−vT+2gT2=gtT+2gT2,T=2(gt+v)2H−gt2=2.996 s,x=vT−2gT2=598.5 ft.
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