Question
A ball is allowed to fall from the top of the tower 200m high. At the same instant another ball is thrown vertically upwards from the bottom of the tower with velocity of 40m/s. When and where the two balls meet?
Answers
ball1:
h=200-4.8t^2
ball2:
h=40*t-4.8t^2
subtract the equations.
0=200-40t
t= 5 seconds
where? h=200-4.8*5^2
h=200-4.8t^2
ball2:
h=40*t-4.8t^2
subtract the equations.
0=200-40t
t= 5 seconds
where? h=200-4.8*5^2
Time=5 seconds & Height=77.5 meters.
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