Question #137059
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?
1
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=gt22.h=\frac{gt^2}{2}.

At the height of the first ball


y=Hh=Hgt22y=H-h=H-\frac{gt^2}{2}

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


Hx=uT+gT22= =T2gh+gT22= =g(tT+T22).H-x=uT+\frac{gT^2}{2}=\\\space\\ =T\sqrt{2gh}+\frac{gT^2}{2}=\\\space\\ =g\bigg(tT+\frac{T^2}{2}\bigg).

The tennis ball will be at the same height at the same time, but its motion will be defined by another equation:


x=vTgT22.x=vT-\frac{gT^2}{2}.

Substitute for x in the above equation:

HvT+gT22=gtT+gT22, T=2Hgt22(gt+v)=2.996 s, x=vTgT22=598.5 ft.H-vT+\frac{gT^2}{2}=gtT+\frac{gT^2}{2},\\\space\\ T=\frac{2H-gt^2}{2(gt+v)}=2.996\text{ s},\\\space\\ x=vT-\frac{gT^2}{2}=598.5\text{ ft}.

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