A ball is dropped from rest at the top of a 100 m tall building. At the same instant, a second ball is thrown upward from the base of the building with an initial velocity of 50 m/s. When will the two balls collide, and at what distance above the street?
The law of motion of the first ball
"y_1=y_0+v_{0y}t-\\frac{gt^2}{2}\\\\\n=100-\\frac{9.8t^2}{2}=100-4.9t^2"The law of motion of the second ball
"y_2=y_0+v_{0y}t-\\frac{gt^2}{2}\\\\\n=50t-\\frac{9.8t^2}{2}=50t-4.9t^2"At the instant when balls collide "y_1=y_2," so
The distance above the street
"d=y_1(t)=y_2(t)=50\\times 2-4.9\\times 2^2=80.4\\:\\rm m"
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